arXiv:1611.06105v1 [] 18 Nov 2016

Distances on a masure (affine ordered hovel)
Auguste Hébert
Univ Lyon, UJM-Saint-Etienne CNRS
UMR 5208 CNRS, F-42023, SAINT-ETIENNE, France
auguste.hebert@univ-st-etienne.fr
Abstract
A masure (a.k.a affine ordered hovel) I is a generalization of the Bruhat-Tits building
that is associated to a split Kac-Moody group G over a non-archimedean local field.
This is a union of affine spaces called apartments. When G is a reductive group, I is
a building and there is a G-invariant distance inducing a norm on each apartment. In
this paper, we study distances on I inducing the affine topology on each apartment. We
show that some properties (completeness, local compactness, ...) cannot be satisfyed
when G is not reductive. Nevertheless, we construct distances such that each element
of G is a continuous automorphism of I.

1

Introduction

If G is a split Kac-Moody group over a non-archimedean local field, Stéphane Gaussent and
Guy Rousseau introduced a space I on which G acts and they called this set a "masure"
(or an "affine ordered hovel"), see [GR08], [Rou12]. We then consider G as a group of automorphisms of I. This construction generalizes the construction of the Bruhat-Tits building
associated to a split reductive group over a local field made by François Bruhat and Jacques
Tits, see [BT72] and [BT84]. This masure is an object similar to a building. It is a union of
subsets called "apartments", each one having a structure of affine space and an additional
structure of apartment defined by hyperplanes (called walls) of this affine space. The group
G acts by permuting the apartments and by inducing affine map on each apartment, sending
walls on walls. We can also define sectors and retractions from I onto apartments with center
a sector, as in the case of Bruhat-Tits buildings. However there can be two points of I which
are not included in any apartment. Studying I enables to get information on G and this is
why we study masures.
Each Bruhat-Tits building B associated to a split reductive group H over a local field is
equipped with a distance d such that H acts isometrically on B and such that the restriction
of d to each apartment is a euclidean distance. This distances are important tools in the
study of buildings. We will show that we cannot equip masures which are not buildings with
distances having this properties but it seems natural to ask whether we can define distances
on a masure having "good" properties. We limit our study to distances inducing the affine
topology on each apartment. We show that under assumptions of continuity for retractions,
the metric space we have is not complete nor locally compact (see Subsection 3.3). We show
that there is no distance on I such that the restriction to each apartment is a norm. However,
we prove the following theorems (Corollary 5.8, Lemma 4.9, Corollary 5.9 and Theorem 5.14):
let q be a sector germ of I, then there exists a distance d on I having the following properties:
1

• the topology induced on each apartment is the affine topology
• each retraction with center q is 1-Lipschitz continuous
• each retraction with center a sector-germ of the same sign as q is Lipschitz continuous
• each g ∈ G is Lipschitz continuous when we see it as an automorphism of I.
We call the distances constructed in the proof of this theorem distances of positive or of
negative type, depending on the sign of q. A distance of positive or negative type is called a
signed distance. We prove that all distances of positive type on a masure (resp. of negative
type) are equivalent, where we say that two distances d1 and d2 are equivalent if there exist
k, l ∈ R∗+ such that kd1 ≤ d2 ≤ ld1 (this is Theorem 5.7). We thus get a positive topology
T+ and a negative topology T− . We prove (Corollary 6.4) that these topologies are different
when I is not a building. When I is a building these topologies are the usual topology on a
building (Proposition 5.15).
Let I0 be the orbit of some special vertex under the action of G. If I is not a building,
I0 is not discrete for T− and T+ . We also prove that if ρ is a retraction centered at a positive
(resp. negative) sector-germ, ρ is not continuous for T− (resp. T+ ), see Proposition 6.3. For
this reasons we introduce combined distances, which are the sum of a distance of positive
type and of a distance of negative type. We then have the following theorem (Theorem 6.7):
all the combined distances on I are equivalent; moreover, if d is a combined distance and I
is equipped with d we have:
• each g : I → I ∈ G is Lipschitz continuous
• each retraction centered at a sector-germ is Lipschitz continuous
• the topology induced on each apartment is the affine topology
• there exists δ > 0 such that for all x, x′ ∈ I0 , d(x, x′ ) < δ implies x = x′ .
The topology Tc associated to combined distances is the initial topology with respect to
the retractions of I (see Corollary 6.10).
Let us explain how we define distances of positive or negative type. Let A be an apartment
and Q be a sector of A. Maybe considering g.A for some g in G, one can suppose that A = A,
the standard apartment of I and Q = Cfv , the fundamental chamber of A (or Q = −Cfv but
this case is similar). Let N be a norm on A. If x ∈ I, there exists an apartment Ax containing
x and +∞ (which means that Ax contains a sub-sector of Cfv ). For q ∈ Cfv , we define x + q
as the translate of x by q in Ax . When q is made more and more dominant, x + q ∈ Cfv .
Therefore, for all x, x′ ∈ I, there exists q, q ′ ∈ Cfv such that x + q = x′ + q ′ . We then define
d(x, x′ ) to be the minimum of the N(q) + N(q ′ ) for such couples q, q ′.
We thus obtain a distance for each sector Q and for each norm N on an apartment
containing Q. We show that this distance only depends on the germ of Q and on N(in
Subsection 4.4) .
In Section 2, we set the general frameworks and define masures.
In Section 3, we show that if q is a sector-germ of I, we can write each apartment as a
finite union of closed convex parts such that each part is included in an apartment containing
q. The most important case for us is when A contains a sector-germ adjacent to q. We can
then write A as the union of two half-apartments, each included in an apartment containing
q. At the end of this part, we show restrictions we have on distances on a masure.
2

In Section 4, we construct signed distances on I. In Section 5, we use the results of the
first section to show that the distances of the same sign are equivalent. We use it to show
the properties mentioned in the abstract. In Section 6, we first show that when I is not
a building, T+ and T− are different. Then we define combined distances and study their
properties. In Section 7, we show that I is contractible for the topologies we introduced.
Acknowledgments I thank Stéphane Gaussent and Guy Rousseau for their comments on
this paper.

2

General frameworks

In this section, we describe masures we are going to study. They have the following properties:
they are semi-discrete, thick of finite thickness and there is a group (vectorially Weyl) acting
strongly transitively on them (see [Rou11] 1 and [GR14] 1 for definitions). This properties
are satisfied by masures associated to (quasi) split Kac-Moody groups over local fields. We
begin by defining the standard apartment. References for this Section are [Kac94], Chapter
1 and 3, [GR08] Section 2 and [GR14] Section 1.

2.1

Root generating system

A Kac-Moody matrix (or generalized Cartan matrix) is a square matrix C = (ci,j )i,j∈I with
integers coefficients, indexed by a finite set I and satisfying:
1. ∀i ∈ I, ci,i = 2
2. ∀(i, j) ∈ I 2 |i 6= j, ci,j ≤ 0
3. ∀(i, j) ∈ I 2 , ci,j = 0 ⇔ cj,i = 0.
A root generating system is a 5-tuple S = (C, X, Y, (αi )i∈I , (αi∨)i∈I ) made of a Kac-Moody
matrix C indexed by I, of two dual free Z-modules X (of characters) and Y (of cocharacters)
of finite rank rk(X), a family (αi )i∈I (of simple roots) in X and a family (αi∨)i∈I (of simple
coroots) in Y . They have to satisfy the following compatibility condition: ci,j = αj (αi∨ ) for
all i, j ∈ I. We also suppose that the family (αi )i∈I is free in X and that the family (αi∨ )i∈I
is free in Y .
We now fix a Kac-Moody matrix C and a root generating system with matrix C.
Let V = Y ⊗ R. Every element of X induces a linear form on V . We will consider X as
a subset of the dual V ∗ of V : the αi , i ∈ I are viewed as linear form on V . For i ∈ I, we
define an involution ri of V by ri (v) = v − αi (v)αi∨ for all v ∈ V . Its space of fixed points is
ker αi . The subgroup of GL(V ) generated by the αi for i ∈ I is denoted by W v and is called
the Weyl group
Lof S.
Let Q∨ = i∈I Zαi∨ . We call Q∨ the coroot-lattice.
One defines an action of the group W v on V ∗ by the following way: if x ∈ V , w ∈ W v
−1
v
and α ∈ V ∗ then (w.α)(x) = α(w
i |(w, i) ∈ W × I}, be the set of real
L .x). Let Φ = +{w.αL
roots. Then Φ ⊂ Q, where Q = i∈I Zαi . Let Q = i∈I Nαi , Q− = −Q+ , Φ+ = Φ ∩ Q+
and Φ− = Φ ∩ Q− . Then Φ = Φ+ ∪ Φ− . The element of Φ+ (resp. Φ− ) are called the real
positive roots (resp. real negative roots). Let W a = Q∨ ⋊ W v ⊂ GA(V ) the affine Weyl group
of S, where GA(V ) is the group of affine isomorphisms of V .

3

A wall M (resp. a half-apartment D) of V is a hyperplane (resp. a half-space) of the
shape α−1 {k} (resp. α−1 ([k, +∞)) for some k ∈ R. The wall M (resp. half-apartment) is
said to be a true wall (resp. a true half-apartment) if k ∈ Z and a ghost wall if k ∈
/ Z.

2.2

Vectorial faces

Define Cfv = {v ∈ V | αi (v) > 0, ∀i ∈ I}. We call it the fundamental chamber. For J ⊂ I,
one sets F v (J) = {v ∈ V | αi (v) = 0 ∀i ∈ J, αi (v) > 0 ∀i ∈ J\I}. Then the closure Cfv of
Cfv is the union of the F v (J) for J ⊂ I. The positive (resp. negative) vectorial faces are
the sets w.F v (J) (resp. −w.F v (J)) for w ∈ W v and J ⊂ I. A vectorial face is either a
positive vectorial face or a negative vectorial face. We call positive chamber (resp. negative)
every cone of the shape w.Cfv for some w ∈ W v (resp. −w.Cfv ). By Section 1.3 of [Rou11],
v
the action
S of W on vthe positive chambers is simply transitive. The Tits cone T v is defined
by T = w∈W v w.Cf . We also consider the negative cone −T . We define a W -invariant
relation ≤ on V by: ∀(x, y) ∈ V 2 , x ≤ y ⇔ y − x ∈ T .

2.3

Study of metric properties of W v

In this subsection we prove that when W v is infinite there exists no norm on V such that
W v is a groupTof isometries T
and we give a result of density of the walls of V .
Let Vin = i∈I ker αi = w∈W v ker(Id − w). Let V = V /Vin . If x ∈ V , one denotes by x
its image in V . Each w ∈ W v induces an automorphism w of V by the formula: w(x) = w(x)
for all x ∈ V . Let W v = {w|w ∈ W v } ⊂ Aut(V ) where Aut(V ) is the group of linear
automorphisms of V . Let Y = {y|y ∈ Y }.
Lemma 2.1.
1. The map f : W v → W v defined by f (w) = w for all w ∈ W v is an
isomorphism of groups.
2. The set Y is a lattice of V .
Proof. Let w ∈ W v \{1} and x ∈ C v . As the action of W v on the positive chambers is simply
transitive, w.x ∈
/ Cfv . Therefore, there exists i ∈ I such that αi (w.x) < 0. In particular
αi (w.x − x) < 0 and thus w.x − x ∈
/ Vin . As a consequence, w 6= 1 and f is injective, which
proves 1.
As Y spans V , Y spans V . Let | | be a norm on V . One equips V with the quotient norm
| |: for all x ∈ V , |x| = inf y∈x |y|.
N
Let (zn ) ∈ Y be a sequence converging towards 0. For all n ∈ N, one writes zn = yn
with yn ∈ Y , one chooses un ∈ yn such that |un | ≤ 2|zn | and one writes un = xn + yn , with
xn ∈ Vin .
T
As V is finite dimensional, there exists w1 , . . . , wl ∈ W v such that Vin = li=1 ker(wi −Id).
For all i ∈ J1, lK, wi .un = xn + wi .yn → 0. Consequently yn − wi yn → 0 for all i ∈ J1, lK. As
yn − wi .yn ∈ Y for all (i, n) ∈ J1, lK × N, yn ∈ Vin for n large enough. Thus zn = yn = 0 for
n large enough, which shows that Y is a lattice of V .
Lemma 2.2. Let Z be a finite dimensional vectorial space on R and H be a subgroup of
GL(Z). Suppose that H is infinite and stabilizes a lattice L. Then for all norm on Z, H is
not a group of isometries.
Proof. Let (e1 , . . . , ek ) be a basis of Z such that L = Ze1 ⊕ . . . ⊕ Zek . Consider f : H → Lk
defined by f (h) = (h.e1 , . . . , h.ek ) for all h ∈ H. Then f is injective and thus there exists
4

i ∈ J1, kK such that {h.ei |h ∈ H} is infinite. As {h.ei |h ∈ H} ⊂ L, {h.ei |h ∈ H} is not
bounded, which shows the lemma.
Let M be a true wall of V , M = β −1 ({k}), with β ∈ Φ and k ∈ Z. If k = 0, one says
that M is a true wall direction. Two true walls M1 = β −1 ({k1 }) and M2 = β −1 ({k2 }) (with
β ∈ Φ and k1 , k2 ∈ Z) of the same directions are called consecutive if |k1 − k2 | = 1.
Proposition 2.3.
1. Suppose that there exists a norm | | on V such that W a is a group
of isometries. Then W v is finite.
2. Let | | be a norm on V , d be the induced distance on V and suppose that W v is infinite.
Then for all ǫ > 0 there exists a true wall direction M0 such that for all consecutive
walls M1 and M2 of direction M0 , d(M1 , M2 ) < ǫ.
Proof. Part 1 is an consequence of Lemma 2.2, applied with Z = V , H = W v and L = Y .
Suppose that W v is infinite. Let | | be a norm on V and suppose that there exists η > 0
such that for all consecutive walls M1 and M2 , d(M1 , M2 ) ≥ η. Let β ∈ Φ and x, y ∈ V such
1
that β(x) 6= β(y). Then | β(y−x)
(y − x)| ≥ η and thus β(y − x) ≤ |y−x|
. For all x ∈ V , one
η
sets N(x) = supβ∈Φ |β(x)|. Define N : V → R+ by N(x) = N(x) for all x ∈ V . Then N is a
norm on V and W v is a group of isometries for this norm. By Lemma 2.1, and Lemma 2.2
applied with Z = V , H = W v and L = Y , this is absurd. Thus the proposition is proved.

2.4

Filters

Definition 2.4. A filter in a set E is a nonempty set F of nonempty subsets of E such that,
for all subsets S, S ′ of E, if S, S ′ ∈ F then S ∩ S ′ ∈ F and, if S ′ ⊂ S, with S ′ ∈ F then
S ∈ F.
If F is a filter in a set E, and E ′ is a subset of E, one says that F contains E ′ if every
element of F contains E ′ . If E ′ is nonempty, the set FE ′ of subsets of E containing E ′ is a
filter. By abuse of language, we will sometimes say that E ′ is a filter by identifying FE ′ and
E ′ . If F is a filter in E, its closure F (resp. its convex envelope) is the filter of subsets of E
containing the closure (resp. the convex envelope) of some element of F . A filter F is said
to be contained in an other filter F ′ : F ⊂ F ′ (resp. in a subset Z in E: F ⊂ Z) if and only
if any set in F ′ (resp. if Z) is in F .
A sector in V is a set of the shape S = x + C v with C v = ±w.Cfv for some x ∈ V and
w ∈ W v . The point x is its base point and C v is its direction. The intersection of two sectors
of the same direction is a sector of the same direction.
The sector-germ of a sector S = x + C v is the filter s of subsets of V containing a V translate of S. It only depends on the direction C v . We denote by +∞ (resp. −∞) the
sector-germ of Cfv (resp. −Cfv ). A sector S = x + C v is said to be positive (resp. negative)
if C v = w.Cfv (resp. C v = −w.Cfv ) for some w ∈ W v . The sign extends to sector-germs.
A ray δ with base point x and containing y 6= x (or the interval ]x, y] = [x, y]\{x} or
[x, y]) is called preordered if x ≤ y or y ≤ x and generic if y − x ∈ ±T̊ , the interior of ±T .
We now denote by A the affine space V equipped with its faces, chimneys, ...
In the next subsection, we give the definition of a masure. This definition uses objects
which we have not defined. This definitions are available in [Rou11] 1.7 and 1.10 and in
[GR14] 1.4. For a first reading, one can just know the following facts about these objects:

5

1. An enclosure clA (F ) is associated to each filter F of A . This is a filter in A containing
the convex envelope of the closure of F .
2. A face or a chimney is a filter in V .
3. A sector is a chimney which is solid and splayed.
4. If a chimney is a sector, its germ as a chimney coincides with its germ as a sector.
5. The group W a permutes the sectors, the enclosures, the faces and the chimneys of V .

2.5

Masure

We now denote by A the affine space V equipped with its faces, chimneys, ...
An apartment of type A is a set A with a nonempty set Isom(A, A) of bijections (called
Weyl isomorphisms) such that if f0 ∈ Isom(A, A) then f ∈ Isom(A, A) if and only if, there
exists w ∈ W a satisfying f = f0 ◦w. A Weyl isomorphism between two apartments φ : A → A′
is a bijection such that (f ∈ Isom(A, A) if and only if φ ◦ f ∈ Isom(A, A′ )). We extend all the
notions that are preserved by W a to each apartment. By the fact 5 of the above subsection,
sectors, enclosures, faces and chimneys are well defined in any apartment of type A.
Definition 2.5. An affine masure of type A is a set I endowed with a covering A of subsets
called apartments such that:
(MA1) Any A ∈ A admits a structure of an apartment of type A.
(MA2) If F is a point, a germ of a preordered interval, a generic ray or a solid chimney
in an apartment A and if A′ is another apartment containing F , then A ∩ A′ contains the
enclosure clA (F ) of F and there exists a Weyl isomorphism from A onto A′ fixing clA (F ).
(MA3) If R is the germ of a splayed chimney and if F is a face or a germ of a solid
chimney, then there exists an apartment that contains R and F .
(MA4) If two apartments A, A′ contain R and F as in (MA3), then there exists a Weyl
isomorphism from A to A′ fixing clA (R ∪ F ).
(MAO) If x, y are two points contained in two apartments A and A′ , and if x ≤A y then
the two segments [x, y]A and [x, y]A′ are equal.
In this definition, we say that an apartment contains a germ of a filter if it contains at
least one element of this germ. We say that a map fixes a germ if it fixes at least one element
of this germ. In the sequel, we will say "isomorphism" instad of "Weyl isomorphism".
Each apartment A can be equipped with a structure of affine space by using an isomorphism of apartments φ : A → A. We say that we fix a vectorial structure on A when we fix
the origin of A at φ(0).
We assume that I has a strongly transitive group of automorphisms G, which means that
all isomorphisms involved in the above axioms are induced by elements of G. We choose
in I a fundamental apartment, that we identify with A. As G is strongly transitive, the
apartments of I are the sets g.A for g ∈ G. The stabilizer N of A induces a group ν(N) of
affine automorphisms of A and we suppose that ν(N) = W v ⋉ Y .
We suppose that I is thick of finite thickness, see [GR14] 1.5 for a definition. This
definition coincides with the usual one when I is a building. We will not use this definition
directly but we will use it to apply results of [Rou11] and finiteness results of [Héb16].
An example of such a masure I is the masure associated to a split Kac-Moody group over
an ultrametric field constructed in [GR08] and in [Rou12].
The masure I is a building if and only if W v is finite, see [Rou11] 2.2 6).
6

2.6

Retractions centered at sector-germs

Let s be a sector-germ of I. Let x ∈ I. By (MA3), there exists an apartment Ax of I
containing x and s. By (MA4), there exists an isomorphism of apartments φ : Ax → A
fixing s. By [Rou11] 2.6, φ(x) does not depend on the choices we made and thus we can let
ρA,R (x) = φ(x).
The map ρA,s is a retraction from I onto A. It only depends on s and A and we call it
the retraction onto A centered at s.
If A and B are two apartments, and φ : A → B is an isomorphism of apartments fixing
X
A∩B
some set X, one writes φ : A → B. If A and B share a sector-germ q, one denotes by A → B
q
or by A → B the unique isomorphism of apartments from A to B fixing q (and also A ∩ B).
q
We denote by I → A the retraction onto A fixing q.

2.7

Parallelism in I and definition of yν and Tν

Let us explain briefly the notion of parallelism in I. This is done more completely in [Rou11]
Section 3.
Let us begin with rays. Let δ and δ ′ be two generic rays in I. By (MA3) and [Rou11]
2.2 3) there exists an apartment A containing sub-rays of δ and δ ′ and we say that δ and
δ ′ are parallel, if these sub-rays are parallel in A. Parallelism is an equivalence relation and
its equivalence classes are called directions. Let Q be a sector of I and A be an apartment
containing Q. One fixes the origin of A in the base point of Q. Let ν ∈ Q and δ = R+ ν.
Then δ is a generic ray in I. By Lemma 3.2 of [Héb16], for all x ∈ I, there exists a unique
ray x + δ of direction δ and base point x. To obtain this ray, one can choose an apartment
Ax containing x and a sub-ray δ ′ of δ, which is possible by (MA3) and [Rou11] 2.2 3), and
then we take the translate of δ ′ in Ax having x as a base point.
Let us recall briefly results from "Definition of yν and Tν " of Section 3 of [Héb16]. Let
x ∈ I, one has x + δ ∩ A = yν (x) + δ for some yν (x) ∈ A. Let q be the germ of Q and
q
ρ : I → A. One has yν (x) = ρ(x) + Tν (x)ν, for some Tν (x) ∈ R+ . This defines two maps
yν : I → A and Tν : I → R+ and we have yν (x) = x and Tν (x) = 0 for all x ∈ A. For all
x ∈ I\A, yν (x) 6= x and Tν (x) > 0.
A sector-face f of A, is a set of the shape x + F v for some vectorial face F v and some
x ∈ A. The germ F = germ∞ (f ) of this sector face is the filter containing the elements of
the shape q + f , for some q ∈ F v . The sector-face f is said to be spherical if F v ∩ T̊ is
nonempty. A sector-panel is a sector-face included in a true wall and spanning this one as an
affine space. A sector-panel is spherical (see [Rou11] 1). We extend these notions to I thanks
to the isomorphisms of apartments. Let us make a summary of the notion of parallelism for
sector-faces. This is also more complete in [Rou11], 3.3.4)).
If f and f ′ are two spherical sector-face, there exists an apartment B containing their
germs F and F′ . One says that f and f ′ are parallel if F = germ∞ (x + F v ) and F′ =
germ∞ (y + F v ) for some x, y ∈ B and for some vectorial face F v of B. Parallelism is an
equivalence relation. The parallelism class of a sector-face germ F is denoted F∞ . If M is
a wall of I, its direction M ∞ is the set of F∞ such that F = germ∞ (f ), with f a spherical
sector-face included in M. We denote by I ∞ the set of directions of spherical faces of I. The
sector-faces of a wall M ∞ of I ∞ are the elements of this wall.
By Proposition 4.7.1) of [Rou11], for all x ∈ I and all F∞ ∈ I ∞ , there exists a unique
sector-face x + F∞ of direction F∞ and with base point x. The existence can be obtained in
the same way as for rays.

7

Let q be a sector-germ of I and x ∈ I. Let A be an apartment containing x and q (such
an apartment exists by (MA3)). Let x + qA be the closure of x + q in A (for the topology
induced by its structure of affine space). Then x + qA does not depend on the choice of A
A∩B
containing x and q: let B be an apartment containing x and q and φ : A → B. Then
A ∩ B = {x ∈ A|φ(x) = x}. By (MA4), φ fixes the enclosure of x + q, which contains x + qA .
Therefore x + qB ⊃ x + qA and by symmetry, x + qA = x + qB . We denote this set by x + q.
Let f, f ′ be sector-faces. One says that f dominates f ′ (resp. f and f ′ are opposite) if
one can write germ∞ (f ) = germ∞ (x + F), germ∞ (f ′ ) = germ∞ (x′ + F ′v ) for some x, x′ ∈ I
v
and F v , F ′v two vectorial faces of a same apartment of I such that F ⊃ F ′v (resp. such that
F ′v = −F v ). By Proposition 3.2 2) and 3) of [Rou11], these notions extend to I ∞ .

3
3.1

Splitting of apartments
Splitting of apartments in two half in two half-apartment

The aim of this section is to show that if A is an apartment, M is a true wall of A, F
is a sector-panel of M ∞ and q is a sector-germ dominating F∞ , there exist two opposite
half-apartments D1 and D2 of A such that their wall is parallel to M and such that for all
i ∈ {1, 2}, Di and q are included in some apartment, which is Lemma 3.6. This section will
enable us to show that for each choice of sign, the distances we construct are equivalent.
The next lemma enables to prove that if the intersection of two apartment is a halfapartment, it is a true half-apartment. In the proof of this lemma, we use the notion of
Hecke paths but we do not define it precisely and we will not use it in the sequel of this
paper. Hecke paths are more or less images by retractions of ordered segments in I. They
are defined in Section 1 of [GR14]. A Hecke path in an apartment A is a piecewise linear
function π : [0, 1] → A satisfying some conditions. The function π is differentiable on [0, 1]
except maybe in a finite number of points and is left differentiable and right differentiable
at each point where it makes sense. We will use the fact that if π is not differentiable in
t ∈ [0, 1], then π(t) is in some true wall.
Lemma 3.1. Let A be an apartment of I on which we fix a structure of vectorial space, Q
be a sector based at 0 and ν ∈ Q. Let yν be as in Subsection 2.7. Let x ∈ I\A. Then yν (x)
is in some true wall of A.
q

Proof. Let B be an apartment containing x and q and φ : B → A. Let τ : [0, 1] → B
defined by τ (t) = φ−1 (φ(x) + 2tTν (x)ν). Then τ is a segment of B such that τ (0) = x and
τ (1) = yν (x) + Tν (x)ν. By Lemma 3.5 a) of [Héb16], x ≤ yν (x). As yν (x) ≤ yν (x) + Tν (x)ν,
−q
τ (0) ≤ τ (1). Let −Q be the sector of A opposite to Q and −q be its germ. Let ρ− : I → A.
By paragraph 2.3 of [GR14], π = ρ− ◦ τ is a Hecke path of shape 2Tν (x) with respect to
′ 1
−Q. For t ∈ [ 12 , 1], τ (t) = π(t) because τ (t) ∈ A. Therefore π+
( 2 ) = 2Tν (x)ν. If t ∈ [0, 21 ),
τ (t) ∈
/ A (by definition of Tν (x)) and thus Lemma 3.6 of [Héb16] applied to τ̃ : t 7→ τ (2t)
′ 1
shows that π−
( 2 ) 6= 2Tν (x)ν. By (iii) of definition of Hecke paths in [GR14], there exists a
real root β such that β(π( 21 )) = β(yν (x)) ∈ Z.
Lemma 3.2. Let A, B be two distinct apartments of I containing a half-apartment D. Then
A ∩ B is a true half-apartment.
Proof. Let us first prove that A ∩ B is a half-apartment. Let M be the wall of D. We fix an
origin of A in M. Let N be a supplementary of M in A included in the tits cone 0 ± T , u be
8

such that N = Ru and such that D ∩ N = R− u. Let n : A → R be such that n(m + λu) = λ
for all m ∈ M and λ ∈ R. Let X = {n(x)|x ∈ A and x + D ⊂ A ∩ B}.
Let λ = sup X and (λn ) ∈ X N be such that λn → λ. For all n ∈ N, λn u + D ⊂ A ∩ B
and as A 6= B, λ < +∞. By (MA2), A ∩ B contains the enclosure of germx ([x, x0 ]\{x}),
where x = λu and x0 = λ0 u and thus A ∩ B ∋ x. Let y ∈ A\(x + D) and suppose y ∈ B.
Let F and F′ be two opposite wall-sector of M and q′Sbe the sector-germ of I dominating F′
and included in D. Then B contains y + F and thus z∈y+F z + q′ = y + D, which is absurd.
Consequently, A ∩ B = x + D: A ∩ B is a half-apartment.
Let F be the wall of x + D. It remains to show that F is true. Let Z = {z ∈
F | z is not included in any true wall except maybe F }. Let µ be a Lebesgue measure on
F . As the set of true walls of A is countable, µ(F \Z) = 0 and thus Z is not empty. Let
z ∈ Z. Let Q be a sector of A ∩ B having z as a base point. One fixes the origin of A in
z. Let ν ∈ Q and yν : I → A be as in the paragraph "Definition of yν and Tν ". Let −ν be
A∩B
the opposite of ν in A. Let φ : A → B and x = φ−1 (−ν). Then x ∈
/ A and yν (x) = z. By
Lemma 3.1, z is in some true wall M and by choice of z we deduce that M = F .
Lemma 3.3. Let M be a true wall of A and φ ∈ W a be an element fixing M. Then
φ ∈ {Id, s}, where s is the reflection of W a with respect to M.
Proof. One writes w = τ ◦ u, with u ∈ W v and τ a translation of A. Then u(M) is a wall
parallel to M. Let M0 be the wall parallel to M containing 0. Then u(M0 ) is a wall parallel
to M0 and containing 0: u(M0 ) = M0 . Let C be a vectorial chamber adjacent to M0 . Then
u(C) is a chamber adjacent to C: u(C) ∈ {C, s0 (C)}, where s0 is the reflection of W with
respect to M0 . Maybe composing u by s0 , one can suppose that u(C) = C and thus u = Id
(because the action of W v on the set of chambers is simply transitive).
If D is a half apartment of I, one sets D, the filter of half apartment containing a
shortening of D. If D1 and D2 are two half-apartments, one says that D1 ∼ D2 if D1 = D2
and one says that D1 and D2 have opposite direction if there exists an apartment A containing
a shortening of them, if their walls are parallel, and if D1 and D2 are not equivalent. One
says say D1 and D2 are opposite if they have opposite directions and if D1 ∩ D2 is a wall.
Lemma 3.4. Let A1 , A2 , A3 be distinct apartments. Suppose that A1 ∩ A2 , A1 ∩ A3 and
A2 ∩ A3 are half-apartments such that A1 ∩ A3 and A2 ∩ A3 have opposite directions. Then
A1 ∩ A2 ∩ A3 = M where M is the wall of A1 ∩ A3 , and for all (i, j, k) ∈ {1, 2, 3}3 such
that {i, j, k} = {1, 2, 3}, Ai ∩ Aj and Ai ∩ Ak are opposite. Moreover, if s : A3 → A3 is the
A ∩A
A ∩A
A ∩A
reflection with respect to M, φ1 : A3 1→ 3 A1 , φ2 : A3 2→ 3 A2 and φ3 : A1 2→ 1 A2 , then
the following diagram is commutative:
A3

s

/

A3

φ2



A2

φ1
φ3

/



A1

Proof. By "Propriété du Y" (Section 4.9 of [Rou11]), A ∩ A1 ∩ A2 is nonempty. Let x ∈
A∩A1 ∩A2 . Let N be the wall parallel to the wall of A1 ∩A3 containing x. Then A1 ∩A2 ∩A3 ⊃
N. Let D1 = A1 ∩ A3 and D2 = A2 ∩ A3 . Then D1 and D2 have opposite directions
and are not disjoint: they contain M and thus A1 ∩ A2 ∩ A3 ⊃ M. Let us show that
φ−1
2 ◦ φ3 ◦ φ1 = s The half-apartments D3 = A1 ∩ A2 and D1 = A1 ∩ A3 contains M
and thus their walls are parallel. Suppose D3 and D2 are equivalent. Then D3 ⊃ D2 and
D1 ∩ D2 = A1 ∩ A2 ∩ A3 = D3 ∩ D2 ⊃ D2 . Therefore, D1 and D2 are equivalent, which
9

is absurd. The half-apartments φ1 (D1 ) = D1 and φ1 (D2 ) have opposite directions in A1 ,
hence φ1 (D2 ) ∼ D3 . Consequently φ3 ◦ φ1 (D2 ) ∼ φ3 (D3 ) = D3 . We also have φ2 (D2 ) = D2 .
−1
Therefore, φ−1
2 ◦ φ3 ◦ φ1 6= Id. By Lemma 3.3, φ2 ◦ φ3 ◦ φ1 = s. As s fixes A1 ∩ A2 ∩ A3 ,
A1 ∩ A2 ∩ A3 ⊂ M and thus A1 ∩ A2 ∩ A3 = M.
Lemma 3.5. Let M ∞ be a wall of I ∞ . Let q be a sector-germ dominating a sector-panel of
direction F∞ ⊂ M ∞ . Let A1 (resp. A2 ) be an apartment containing a wall M1 (resp. M2 ) of
direction M ∞ and q. Then either A1 = A2 or A1 ∩ A2 is a half-apartment.
Proof. Let Q be a sector of germ q and included in A1 ∩ A2 . Let N1 be a wall parallel to
A ∩A
M1 and such that N1 ∩ Q 6= ∅. Let φ : A1 1→ 2 A2 and N2 = φ(N1 ). Let f ⊂ N1 ∩ B be a
sector-panel of direction F∞ . Then N2 ⊃ f . Let H be a wall of A2 parallel to M2 such that
H ⊃ f . Then H is the affine space of A2 spanned by f and N2 too: H = N2 . Therefore,
N2∞ = M ∞ = N1∞ . By Proposition 4.8 2) of [Rou11] N1 = N2 (this is the unique wall of
direction M ∞ containing F = germ∞ (f )). Let F′ be the sector-panel germ of N1 opposite to
F = germ∞ (f ). Then by (MA4), A1 ∩ A2 ⊃ cl(F′ ∪ q), which is a half apartment and one
concludes with Lemma 3.2.
The following lemma is similar to Proposition 2.9.1) of [Rou11].
Lemma 3.6. Let A be an apartment, M be a wall of A and M ∞ be its direction. Let F∞ be
the direction of a sector-panel of M ∞ and q be a sector-germ dominating F∞ and not included
in A. Then there exists a unique pair {D1 , D2 } of half-apartments of A such that:
• D1 and D2 are opposite with common wall N parallel to M
• for all i ∈ {1, 2}, Di and q are in some apartment Ai .
Moreover:
• D1 and D2 are true half-apartments
• such apartments A1 and A2 are unique and if D is the half-apartment of A1 opposite
to D1 , then D ∩ D2 = D1 ∩ D2 is a wall and A2 = D2 ∪ D.
Proof. Let us first show the existence of D1 and D2 . Let F′∞ be the sector-panel of M ∞
opposite to F∞ . Let q′1 and q′2 be the sector-germ of A containing F′∞ . For i ∈ {1, 2}, let Ai
be an apartment
of I containing q′i and q. Let i ∈ {1, 2} and x ∈ A∩Ai . Then x+q′i ⊂ A∩Ai
S
and Ei = y∈x+q′ y + F∞ ⊂ A ∩ Ai is a half-apartment of A and Ai .
i
S
Suppose A1 = A2 . Then A1 ⊃ x∈E1 x + q′2 = A and thus A1 = A ⊃ q, which is absurd.
For i ∈ {1, 2}, Ei and thus Ai contains a wall of direction M ∞ . By Lemma 3.5, A1 ∩ A2 is a
half-apartment.
By Lemma 3.4, A1 ∩ A2 ∩ A = N, where N is a wall of A parallel to M, and if Di = A ∩ Ai
for all i ∈ {1, 2}, {D1 , D2 } fulfills the condition of the lemma. Let E1 , E2 be another pair
of opposite half apartments of A such that for all i ∈ {1, 2}, Ei and q are included in some
apartment Bi and such that E1 ∩E2 is parallel to M. One can suppose E1 ∼ D1 and E2 ∼ D2 .
Suppose for example that E1 ⊂ D1 . By the same reasoning as in the proof of the existence
of D1 and D2 , B1 ∩ A2 is a half apartment. Thus by Lemma 3.4, B1 ∩ A2 ∩ A is a wall of
A. But B1 ∩ A2 ∩ A = E1 ∩ D2 ⊂ D1 ∩ D2 = N, thus E1 ∩ D2 = N and hence E1 ⊃ D1 .
Therefore, E1 = D1 and by symmetry, E2 = D2 . This proves the uniqueness of such a pair.
The fact that D1 and D2 are true half-apartments comes from Lemma 3.2. Let us show the
uniqueness of such apartments A1 and A2 . Let C1 be an apartment containing D1 and q.
Let F be the sector-panel germ of N dominated by q. Let F′ be the sector-panel germ of N
opposite to F. Then by (MA4), C1 ⊃ cl(F′ , q) = D and thus C1 ⊃ A1 . Therefore, A1 = C1 .
By symmetry, we get the lemma.
10

3.2

Splitting of apartments

In this subsection we mainly generalize Lemma 3.6. We show that if q is a sector germ of I
and if A is an apartment of I, then A is the union of a finite number of convex closed parts
Pi of A such that for all i, Pi and q are included in some apartment. This is Proposition 3.7.
We then deduce a property of retractions with respect to a sector-germ. This subsection will
not be used in the study of the distances we construct.
From now on, unless otherwise stated, "half-apartment" will implicitly refer to "true
half-apartment".
Let q be a sector-germ and ǫ ∈ {+, −} be its sign. Let A be an apartment of I. Then
one sets dq (A) = min{d(q, q′)|q′ is a sector germ of A of sign ǫ}. LetTDA be the set of halfapartments of A. One sets PA,0 = {A} and for all n ∈ N∗ , PA,n = { ni=1 Di |(Di ) ∈ (DA )n }.
The following proposition is very similar to Proposition 4.3.1 of [Cha10].
Proposition 3.7. Let A be an apartment of I, q be a sector-germ of I et n = dq (A). Then
S
there exist P1 , . . . , Pk ∈ PA,n , with k ≤ 2n such that A = ki=1 Pi and for each i ∈ J1, kK,
Pi and q are contained in some apartment Ai . Moreover, for all i ∈ J1, kK, there exists an
P
isomorphism ψi : Ai →i A.
Proof. We do it by induction on n. This is clear if n = 0. Suppose this is true for all
apartment B such that dq (B) ≤ n − 1. Let s be a sector-germ of A such that there exists a
minimal gallery s = q1 , . . . , qn = q from q to s. Let M be a wall of A containing a sectorpanel F dominated by q1 and q2 . Let D1 , D2 be a pair of opposite half-apartments of wall
parallel to M and such that for all i ∈ {1, 2}, Di , q2 is included in an apartment Bi (such a
Si
(i)
Pj ,
pair exists by Lemma 3.6). Let i ∈ {1, 2}. One has dq (Bi ) = n − 1 and thus Bi = kj=1
(i)

(i)

with ki ≤ 2n−1 , for all j ∈ J1, ki K, Pj ∈ DBi ,n−1 and q, Pj
(i)
Aj . One has
[
A = D1 ∩ B1 ∪ D2 ∩ B2 =

is contained in some apartment
(i)

P j ∩ Di .

i∈{1,2},j∈J1,ki K

A∩B

(i)

(i)

Let i ∈ {1, 2} and φi : Bi → i A. Let j ∈ J1, ki K. Then we still have Pj ∩ Di ⊂ Aj . One
Tn−1
(i)
(i)
(i)
(i)
n−1
El with (El ) ∈ DB
writes Pj = l=1
. We have Pj ∩ Di = φi (Pj ∩ Di ) = φi (Pj ) ∩ Di ,
i
and thus
n−1
\
(i)
φi (El ) ∈ PA,n .
Pj ∩ Di = Di ∩
l=1

(j)

(j) Pi

This shows the first part of this lemma. Let i ∈ {1, 2} and j ∈ J1, ki K. Let f : Ai

→ Bi .

(i)

Let ψ = φi ◦ f . One has ψ|P (i) ∩Di = φi ◦ IdP (i) ∩Di = IdP (i) ∩Di . Thus ψ : A
j
j
j
completes the proof.

Pj ∩Di

→

A, which

If A is an apartment and x, y ∈ A, one denotes by [x, y]A the segment joining x and y in
A.
We deduce from the previous proposition a corollary which was already known for hovels
associated to split Kac-Moody groups over ultrametric fields by Section 4.4 of [GR08]:
Corollary 3.8. Let q be a sector-germ, ASbe an apartment and x, y ∈ A. Then there exists
k−1
[xi , xi+1 ]A and such that for all i ∈ J1, k − 1K,
x = x1 , . . . , xk = y ∈ A such that [x, y]A = i=1
[xi , xi+1 ]A and q are included in some apartment.
11

Corollary 3.9. Let A, B be two apartments of I, q be a sector-germ of A and ρ : B → A be
the retraction on A with center q. We choose norms on A and B. Then if a subset X of B
is of nonempty interior, ρ(X) is of nonempty interior.
S
Proof. Let n = dq (B). One chooses k ≤ 2n and P1 , . . . , Pk ∈ PB,n such that B = ki=1 Pi and
such that for all i ∈ J1, kK, Pi , q are in some apartment Ai . For all i ∈ J1, kK, one chooses
P
ψi : Ai →i B. Let X ⊂ B be a set with nonempty interior. Then there exists i ∈ J1, kK such
that Pi ∩ X is of nonempty interior in B. Then X ∩ Pi = ψi (Pi ∩ X) is of nonempty interior
q
in Ai . Let φ : Ai → A. Then ρ(X ∩ Pi ) = φ(Pi ∩ X) is of nonempty interior.

3.3

Restrictions on the distances

In this section, we show that some properties cannot be satisfied by distances on masures. If
A is an apartment of I, we show that there exist apartments branching at all wall of A (this
is Lemma 3.10). This implies that if I is not a building the interior of each apartments is
empty for the distances we study. We write I as a countable union of apartment and then
use Baire’s Theorem to show that under rather weak assumption of regularity for retractions,
a masure cannot be complete or locally compact for the distances we study.
Let us show a slight refinement of Corollaire 2.10 of [Rou11]:
Lemma 3.10. Let A be an apartment of I and D be a half-apartment of A. Then there
exists an apartment B such that A ∩ B = D.
Proof. By Corollaire 2.10 of [Rou11], D is the intersection of two apartments, say A1 and A2 .
If A1 = A or A2 = A, there is nothing to prove. Suppose A1 , A2 6= A. Suppose A1 ∩ A 6= D
and A2 ∩ A 6= D. Then by Lemma 3.2, for all i ∈ {1, 2}, Ai ∩ A = Di , where Di is a half
apartment of A containing strictly D. But then A1 ∩ A2 ⊃ D1 ∩ D2 ) D, which is absurd.
Therefore A1 ∩ A = D or A2 ∩ A = D.
Proposition 3.11. Suppose that there exists a distance dI on I such that for all apartment
A, dI|A2 is induced by some norm. Then I is a building and dI|A2 is W a -invariant.
q

Proof. Let q be a sector germ and A, B be two apartments containing q. Let φ : A → B.
Let us show that φ : (A, dI ) → (B, dI ) is an isometry. Let d′ : A × A → R+ defined
by d′ (x, y) = dI (φ(x), φ(y)) for all x, y ∈ A. Then d′ is induced by some norm. Therefore
d′|(A∩B)2 = dI|(A∩B)2 and as A∩B contains an open set, d′ = dI and thus φ : (A, dI ) → (B, dI )
is an isometry.
Let M be a wall of A, D1 and D2 be the half-apartments defined by M and s ∈ W a
be the reflection with respect to M. Let B be an apartment of I such that A ∩ B = D,
which exists by Lemma 3.10. Let D3 be the half-apartment of A1 opposite to D1 . Then
D3 ∩ D2 ⊂ D3 ∩ A ⊂ M and thus D2 ∩ D3 = M. By Proposition 2.9 2) of [Rou11], D3 ∪ D2
A∩A
A∩A
A ∩A
is an apartment A2 of I. Let φ1 : A → 1 A1 , φ2 : A → 2 A2 and φ3 : A1 2→ 1 A2 . Then by
Lemma 3.4, the following diagram is commutative:
A

s

/

A .

φ2



A2

φ1
φ3

/



A1

Therefore s is an isometry and thus W a is a group of isometries for dI|A2 . By Proposition 2.3 1, W v is finite and by [Rou11] 2.2 6), I is a building.
12

Lemma 3.12. Let q be a sector-germ of I and d be a distance on I inducing the affine
topology on each apartment and such that there exists a continuous retraction of I centered
at q. Then each apartment containing q is closed.
q

q

Proof. Let A be an apartment containing q and B = ρ(I). Let φ : B → A and ρA : I → A.
Then ρA = φ ◦ ρ is continuous because φ is an affine map. Let (xn ) ∈ AN be a converging
sequence and x = lim xn . Then xn = ρA (xn ) → ρA (x) and thus x = ρ(x) ∈ A.
Proposition 3.13. Suppose I is not a building. Let d be a distance on I inducing the affine
topology on each apartment. Then the interior of each apartment of I is empty.
Proof. Let U be a nonempty open set of I. Let A be an apartment of I such that A ∩ U 6= ∅.
By Proposition 2.3 2, there exists a true wall M of A such that D ∩ U 6= ∅. Let D be a
half-apartment D delimited by M. Let B be an apartment such that A∩B = D, which exists
by Lemma 3.10. Then B ∩ U is an open set of B containing M ∩ U and thus E ∩ U 6= ∅,
where E is the half-apartment of B opposite to D. Therefore U\A 6= ∅ and we get the
proposition.
One sets I0 = G.0 where 0 ∈ A. This is the set of vertices of type 0. Let −∞ =
−∞
germ∞ (−Cfv ). One sets ρ−∞ : I → A.
Recall the definition of the lattice Y in Subsection 2.1).
Lemma 3.14. One has I0 ∩ A = Y .
Proof. Let x ∈ I0 ∩ A. One has x = g.0 with g ∈ G. By (MA2), there exists φ : g.A → A
fixing x. Then x = φ(g.x) and φ ◦ g|A : A → A is an automorphism of apartment and hence
x ∈ Y (by the end of Subsection 2.5).
Lemma 3.15. The set I0 is countable.
Proof. Let i ∈ {−∞, +∞}. BySdefinition of ρi , ρi (x) ∈ I0 for all x ∈ I0 and thus ρi (x) ∈ Y
−1
for all x ∈ I0 . Therefore I0 = (λ,µ)∈Y 2 ρ−1
−∞ ({λ}) ∩ ρ+∞ ({µ}). By Theorem 5.6 of [Héb16],
−1
2
ρ−1
−∞ ({λ}) ∩ ρ+∞ ({µ}) is finite for all (λ, µ) ∈ Y , which completes the proof.
Let q be a sector-germ of I. For all z ∈ I0 , one chooses an apartment A(z) containing z
and q. Let x ∈ I and A be an apartment containing
S x and q. There exists z ∈ I0 ∩ A such
that x ∈ z + q and thus x ∈ A(z). Therefore I = z∈I0 A(z).
Proposition 3.16. Let d be a distance on I. Suppose that there exists a sector-germ q
such that each apartment containing q is closed and with empty interior. Then (I, d) is not
complete and the interior of each compact of I is empty.
S
Proof. One has I = z∈I0 A(z), with I0 countable by Lemma 3.15. Thus by Baire’s Theorem,
(I, d) is not complete.
Let K be aScompact of I and U ⊂ K be open. Then U is compact and thus complete.
One has U = z∈I0 U ∩ A(z) and thus U has empty interior. Thus K has empty interior.

13

4
4.1

Construction of signed distances
Definition of positive and negative distances

In this section we construct distances on I. To each sector-germ q and to each norm on an
apartment containing q, we associate a distance on I. Let us be more precise.
Let A be an apartment of I. Let d be a distance on A. One says that d is a norm if there
exists an isomorphism φ : A → A and a norm | | on A such that d(x, y) = |φ−1(x) − φ−1 (y)|
for all (x, y) ∈ A2 . A distance d on A is a norm if and only if for all isomorphism φ : A → A,
there exists a norm | | on A such that d(x, y) = |φ−1(x) − φ−1 (y)| for all x, y ∈ A. Let N (A)
be the set of norms on A. Let q be a sector-germ of I and A(q) be the set of apartments
containing q. Let A ∈ A(q) and d ∈ N (A). Then one sets θq (d) = (dB )B∈A(q) , where
A∩B

−1
−1
for all B ∈ A(q) and
S all x, y ∈ B, dB (x, y) = d(φ (x), φ (y)), where φ : B → A. Let
Θ(q) = {θq (d)| d ∈ A∈A(q) N (A)}. In this section, we fix q and we associate to each θ ∈ Θ(q)
a distance dθ on I.

4.2

Translation in a direction

In this subsection we define for all sector Q of direction q a map + : I × Q such that for all
x ∈ I and q ∈ Q, x + q is the "translate of x by q".
Let A ∈ A(q). One chooses an isomorphism φ : A → A. We consider A as a vectorial
q
space over R via this isomorphism. Let Q = 0 + q ⊂ A. Let ρ : I → A.
Lemma 4.1. Let x ∈ I. Then ρx : x + q → ρ(x) + q is well defined and is a bijection.
q

Proof. Let B be an apartment containing x + q and φ : B → A. Then φ(x + q) = φ(x) + q =
ρ(x) + q. Let u, u′ ∈ x + q be such that ρ(u) = ρ(u′ ). Then u, u′ ∈ B, thus φ(u) = φ(u′) =
ρ(u) = ρ(u′ ) and thus u = u′ .
Let x ∈ I and u ∈ Q. One sets x + u = ρ−1
x (ρ(x) + u). If B is an apartment containing
x and q, then for all u ∈ Q, x + u ∈ B (because B ⊃ x + q).
Lemma 4.2. Let x ∈ I. Then for all u, v ∈ Q, (x+u)+v = x+(u+v) and x+u+v = x+v+u.
A∩B

Proof. Let B be an apartment containing q and x. Let φ : A → B. This isomorphism
enables to consider B as an affine space under the action of A. For u ∈ Q, one denotes by
τu : B → B the translation of vector u. Then for all u ∈ Q, τu (x) = x + u, which proves the
lemma.

4.3

Definition of a distance

Let θ = (dB )B∈A(q) ∈ Θ(q). For all u ∈ A, one sets |u| = dA (u, 0). For x, y ∈ I, one sets
2
T (x, y) = {(u, v) ∈ Q |x + u = y + v}. One defines dθ (x, y) = inf (u,v)∈T (x,y) |u| + |v|. Until
the end of thisTsection, we will write d instead of dθ .
Let Ain = i∈I ker αi ⊂ A (where the αi are the simple roots of A). Let Ain = φ(Ain ).
Lemma 4.3. Let a, b ∈ I. Then T (a, b) is not empty.
Proof. Let (ei )i∈J be a L
basis of A such that for some J ′ ⊂ J, (ei )i∈J\J ′ is a basis of Ain and
such that Q = Ain ⊕ i∈J ′ R∗+ ei . Let (e∗i ) be the dual basis of (ei ). Let B ∈ A(q) and
14

q

φ : A → B. For i ∈ J, one defines e∗i,B : B → R by e∗i,B (x) = e∗i (φ−1 (x)) for all x ∈ B. Then
for i ∈ J, x ∈ B and q ∈ Q, e∗i,B (x + q) = e∗i,B (x) + e∗i (q).
For x ∈ I one chooses Bx ∈ A(q) containing x. Let Qx ⊂ Q be a sector of direction
q included in A ∩ Bx and Mx ∈ R such that for all y ∈ A, mini∈J ′ e∗i (y) ≥ Mx implies
y ∈ Qx ⊂ A ∩ Bx . One chooses qx ∈ Q such that mini∈J ′ (e∗i,Bx (x) + e∗i (qx )) ≥ Mx . Then
x + qx ∈ A for all x ∈ I. Therefore a + qa and b + qb are in A. Thus there exists r, s ∈ Q
such that a + qa + r = b + qb + s and (qa + r, qb + s) ∈ T (a, b).
Proposition 4.4. The function d : I 2 → R+ is a distance.
Proof. The function d is clearly symmetric. Let us show the triangular inequality. Let
x, y, z ∈ I. Let ǫ > 0 and (u, v) ∈ T (x, y), (µ, ν) ∈ T (y, z) be such that |u| + |v| ≤ d(x, y) + ǫ
and |µ| + |ν| ≤ d(y, z) + ǫ. One has x + u = y + v and y + µ = z + ν. Thus x + u + µ =
y + µ + v = z + ν + v (by Lemma 4.2) and hence (u + µ, ν + v) ∈ T (x, z). Consequently,
d(x, z) ≤ |u + µ| + |ν + v| ≤ |u| + |v| + |µ| + |ν| ≤ d(x, y) + d(y, z) + 2ǫ, which proves the
triangular inequality. Let x, y ∈ I be such that d(x, y) = 0. Let (un , vn ) ∈ T (x, y)N be such
that u →
S 0 and v → 0. Let n ∈ N. One has y + q ⊃ y + vn + q = x + un + q and thus
y + q ⊃ n∈N x + un + q = x + q. By symmetry, x + q ⊃ y + q and hence x + q = y + q. Let B
be an apartment containing x and q. By (MA2), B ⊃ cl(x + q) = cl(y + q) ∋ y. Therefore,
x = y.
One equips I × Q with a distance d defined by d((x, q), (x′ , q ′ )) = d(x, x′ ) + |q − q ′ |.
Lemma 4.5. The map

I ×Q→I
is Lipschitz continuous.
(x, ν) 7→ x + ν

Proof. By the fact that each norm are equivalent on an affine space of finite dimension, one
′
can choose a particular θ ∈ Θ(q). Let (e
Li )i∈J be ∗a basis of ∗A such that for some J ⊂ J,
(ei )i∈J\J ′ is a basis of Ain and Q = Ain ⊕ i∈J\J ′ R+ ei and (ei ) be the dual basis to (ei ). For
P
x ∈ A, one sets |x| = i∈I |e∗i (x)| and one supposes that θ is associated to | |. Let us show
that + is 1-Lipschitz continuous.
Let x, x′ ∈ I and ǫ > 0. Let (u, u′) ∈ T (x, x′ ) such that d(x, x′ ) ≤ |u| + |u′ | + ǫ.
∈ J. Let U = {i ∈ J| λi > 0} and
Let ν, ν ′ ∈ Q. One sets λi =Pe∗i (ν − ν ′ ) for all i P
V = {i ∈ J| λi < 0}. Let µ′ = i∈U λi ei and µ = − i∈V λi ei . One has ν + µ = ν ′ + µ′ and
|µ| + |µ′| = d(ν, ν ′ ). One has x + ν + u + µ = x′ + ν ′ + u′ + µ′ and thus d(x + ν, x′ + ν ′ ) ≤
|u| + |u′| + |µ| + |µ′ | ≤ d(x, x′ ) + d(ν, ν ′ ) + ǫ, which enables to conclude.
Remark 4.6. A consequence of Lemma 4.5 is the fact that for all x, y ∈ Q, d(x, y) =
minu,v∈T (x,y) |u| + |v|.
Proposition 4.7. For all x, y ∈ I, there exists a geodesic from x to y.
Proof. Let x, y ∈ I and (u, v) ∈ T (x, y) such that d(x, y) = |u|+|v|. One defines τ : [0, 1] → I
by τ (t) = x + 2tu if t ∈ [0, 12 ] and τ (t) = y + 2(1 − t)v if t ∈ [ 21 , 1], and τ is a geodesic from x
to y.
Remark 4.8. If dim A ≥ 2, for any choice of θ, there exists two points in I such that there
exist infinitely many geodesics between them. For example, if we choose the norm
P on A as in
the proof of Lemma 4.5, we have that d|A2 is the distance induced by | |. If x = i∈I xi ei ∈ A
and if for all i ∈ I, fi : [0, 1] → R is a continuous monotonic function such that fi (0) = 0
and fi (1) = xi , then f = (fi )i∈I is a geodesic from 0 to x.
15

Lemma 4.9. Let B and C be two apartments containing q. Then:
q

(i) the retraction ρB : (I, d) → (B, d|B2 ) is 1-Lipschitz continuous.
q

(ii) the isomorphism ϕ : (B, d|B2 ) → (C, d|C 2 ) is an isometry.
q

Proof. Let ρ : I → A. Let x, y ∈ I and (u, v) ∈ T (x, y). One has x + u = y + v,thus
ρ(x + u) = ρ(x) + u = ρ(y + v) = ρ(y) + v and hence (u, v) ∈ T (ρ(x), ρ(y)), which proves
that ρ is 1-Lipschitz continuous.
First suppose that C = A. Let x, y ∈ B and (u, v) ∈ T (ρ(x), ρ(y)). Then x + u =
−1
ρx (ρ(x) + u) = ϕ−1 (ρ(x) + u) = ϕ−1 (ρ(y) + v) = y + v. Hence T (x, y) ⊃ T (ρ(x), ρ(y))
and thus T (x, y) = T (ρ(x), ρ(y)). Therefore, ϕ is an isometry. Suppose now C 6= A. Let
q
q
q
ϕ1 : B → A, ϕ2 : A → C and ϕ : B → C. Then ϕ = ϕ2 ◦ ϕ1 is an isometry. One has
ρB = ϕ−1
1 ◦ ρ and thus ρB is 1-Lipschitz continuous.

4.4

Independence of the choices of apartments and isomorphisms

Let us show that the distance we defined only depends on θ ∈ Θ(q). For this we have to
show that d is independent of the choice of A ∈ A(q) and of the isomorphism φ : A → A.
q
Let A′ ∈ A(q), f : A → A′ and φ′ = f ◦ φ. One considers A′ as a vectorial space over R by
saying that f is an isomorphism of vectorial space. Objects or operations in A′ are denoted
with a ′ .
Lemma 4.10. Let x ∈ I and u ∈ Q. Then x + u = x +′ f (u).
Proof. One has ρ′ (x+u) = f ◦ρ(x+u) = f (ρ(x)+u) = f (ρ(x))+′ f (u) = ρ′ (x)+′ f (u). Let B
be an apartment containing x and q. Then B ∋ x+u, x+′ f (u) and thus x+u = x+′ f (u).
Lemma 4.11. One has d = d′ .
Proof. Let x, y ∈ I. By Lemma 4.10, TA′ (x, y) = f (TA (x, y)). Let u ∈ A. Then |f (u)′ | =
dA′ (0′ , f (u)) = dA′ (f (0), f (u)) = dA (0, u) = |u| by definition of Θ(q). Therefore, d = d′ .
Let now ψ : A → A be an other isomorphism of apartments. This defines an other
structure of vectorial space on A. We put a subscript ψ or φ to make the difference between
these structures.
Lemma 4.12. One has dφ = dψ .
Proof. One has ψ = φ ◦ w with w ∈ W . One writes w = τ ◦ w,
~ where w
~ is the vectorial part
−1
−1
of w and τ is a translation of A. Let us show that if τ̃ = φ ◦ τ ◦ φ , x +ψ u = x +φ τ̃ (u)
for all (x, u) ∈ I × Qψ . Let x, u ∈ A. One has x +ψ u = ψ(ψ −1 (x) + ψ −1 (u)) = φ φ−1 (x) +


w(w
~ −1 ◦ φ−1 (u)) = φ φ−1 (x) + φ−1 (τ̃ (x)) = x +φ τ̃ (u). Therefore, for all (x, u) ∈ I × Qψ ,
−1
x +ψ u = ρ−1
x (x +ψ u) = ρx (x +φ τ̃ (u)) = x +φ τ̃ (u). Let us show that |u|ψ = |τ̃ (u)|φ for all
u ∈ Qψ . Let x ∈ A. Then x +ψ 0ψ = x +φ τ̃ (0ψ ) = x +φ 0φ and thus 0φ = τ̃ (0ψ ). Let u ∈ A.
Then |u|ψ = d(u, 0ψ ) = d(τ̃ (u), τ̃(0ψ )) = d(τ̃ (u), 0φ ) = |τ̃ (u)|φ because τ̃ is a translation of
A. Let x, y ∈ I. Then Tψ (x, y) = τ̃ (Tφ (x, y)) and thus dψ (x, y) = dφ (x, y).
Thus we have constructed a distance dθ for all θ ∈ Θ(q). When I is a tree, we obtain the
usual distance.

16

5

Comparison of distances of the same sign

The aim of this section is to show that if q and q′ are sector-germs of I of the same sign and
if θ ∈ Θ(q), θ′ ∈ Θ(q′ ) then dθ and dθ′ are equivalent, which means that there exists k, l ∈ R∗+
such that kdθ ≤ dθ′ ≤ ldθ . To prove this we make an induction on the distance between q
and q′ . In the next subsection, we treat the case where q and q′ are adjacent. We use results
of Subsection 3.1.
Lemma 5.1. Let q be a sector germ of I. Then for all θ ∈ Θ(q) and all A ∈ A(q),
dθ|A2 ∈ N (A)
Proof. Let A ∈ A(q) and θ ∈ Θ(q). One equips A with a structure of vectorial space. Let N :
A → R+ defined by N(x) = dθ (x, 0) for all x ∈ A. Let x, y ∈ A. Then T (x, y) = T (0, y − x)
and thus dθ (x, y) = N(y − x). Let λ ∈ R∗ and x ∈ A. Then T (0, λx) = |λ|T (0, x) and thus
N(λx) = |λ|N(x).

5.1

Comparison of distance for adjacent sector-germs

Let A be an apartment of I and q, q′ be two adjacent sector-germs of A. Let d be a norm
on A, θ = θq (d) and θ′ = θq′ (d). Let dq = dθ and dq′ = dθ′ . We fix a vectorial structure on
A. One sets |x| = d(0, x) for all x ∈ A. Let Q = 0 + q and Q′ = 0 + q′ . For all x, y ∈ I and
R ∈ {Q, Q′ }, one sets TR (x, y) = {u, v ∈ R|x + u = y + v}. The aim of this subsection is to
q

q′

show that there exists k ∈ R such that dq ≤ kdq′ . Let ρq : I → A and ρq′ : I → A.
Lemma 5.2. There exists l ∈ R∗+ such that for all B ∈ A(q)∩A(q′ ) we have: for all x, y ∈ B,
dq′ (x, y) ≤ ldq (x, y).
Proof. Let x, y ∈ B. By Lemma 4.9, dq (x, y) = dq (ρq (x), ρq (y)) and dq′ (ρq′ (x), ρq′ (y)) =
dq′ (x, y). As ρq′ |B = ρq|B , one can suppose that x, y ∈ A. Then this is a consequence of
Lemma 5.1.
Let B be an apartment containing q but not q′ . Let F∞ be the direction of sector-panel
common to q and q′ . Let x ∈ B and N be a wall containing x + F∞ . By Lemma 3.6, one can
write B = D1 ∪ D2 , with D1 and D2 two opposite half-apartments having a wall H parallel to
N, such that Di and q′ are included in some apartment Bi , for all i ∈ {1, 2}. One supposes
that D1 ⊃ q.
Let M be the wall of A containing 0 + F∞ , t0 : A → A be the reflection with respect to
this wall and T ∈ R+ such that t0 : (A, dq ) → (A, dq ) is T -Lipschitz continuous (such a T
exists by Lemma 5.1). As t0 is an involution, T ≥ 1.
Lemma 5.3. There exists a translation τ of A such that if t̃ = τ ◦ t0 , one has for all x ∈ B:
• if x ∈ D1 , ρq (x) = ρq′ (x)
• if x ∈ D2 , ρq (x) = t̃ ◦ ρq′ (x)
B ∩B

B∩B

Proof. Let φi : B → i Bi , for i ∈ {1, 2} and φ : B2 1→ 2 B1 . Let t be the reflection of B1
with respect to H. By Lemma 3.4, one has the following commutative diagram:
B

φ2

/

B2 .

φ1



B1

φ
t

17

/



B1

B ∩A

1
Let x ∈ D1 . Then ρq,B1 (x) = x = ρq′ ,B1 (x). Let ψ : B1 →
A. Then ρq (x) = ψ(ρq,B1 (x)) =
ψ(ρq′ ,B1 (x)) = ρq′ (x).
Let x ∈ D2 . One has ρq,B1 (x) = φ1 (x) and ρq′ ,B1 (x) = φ(x) and thus ρq,B1 (x) = t◦ρq′ ,B1 (x).
Let t̃ making the following diagram commute:

B1

t

B1 .
/

ψ

ψ



A

t̃

/



A

Then ρq (x) = t̃◦ρq′ (x). Moreover t̃ fixes ψ(H), which is parallel to M. Therefore, t̃ = τ ◦t0
for some translation τ of A (by Lemma 3.3).
Lemma 5.4. Let l be as in Lemma 5.2. Let x, y ∈ B be such that x, y ∈ Di for some
i ∈ {1, 2}. Then dq′ (x, y) ≤ lT dq (x, y).
Proof. By Lemma 4.9, dq (x, y) = dq (ρq (x), ρq (y)) and dq′ (x, y) = dq′ (ρq′ (x), ρq′ (y)). Lemma 5.3
completes the proof.
Lemma 5.5. Let (X, d) be a metric space, f : (I, dq ) → (X, d) be a map and k ∈ R+ . Then
f is k-Lipschitz continuous if and only if for all apartment A containing q, f|A is k-Lipschitz
continuous.
Proof. One implication is clear. Suppose that for all apartment A containing q, f|A is kLipschitz continuous. Let x, y ∈ I. Let (u, v) ∈ T (x, y) such that |u| + |v| = d(x, y). One
has d(f (x), f (y)) ≤ d(f (x), f (x + u)) + d(f (y + v), f (y)) ≤ k(|u| + |v|) ≤ kd(x, y).
Lemma 5.6. One has dq′ ≤ lT dq .
Proof. Let us prove that Id : (I, dq ) → (I, dq′ ) is lT -Lipschitz continuous. Let x, y ∈ B.
We already know that if x, y ∈ Di for some i ∈ {1, 2}, dq′ (x, y) ≤ lT dq (x, y). Suppose now
x ∈ D1 and y ∈ D2 . Let u ∈ [x, y] ∩ H. By Lemma 5.1, one has:
dq′ (x, y) ≤ dq′ (x, u) + dq′ (u, y) ≤ lT (dq (x, u) + dq (u, y)) = lT dq (x, y).
As B is an arbitrary apartment containing q and not q′ , Lemma 5.2 and Lemma 5.5 completes
the proof.

5.2

Comparison of distances for sector-germs of the same signs

In this subsection, we show that if q and q′ are sector-germs of the same signs, dθ and dθ′ are
equivalent. We then deduce corollaries about the induced topologies on each apartment and
on retractions centered at a sector-germ.
Let Θ be the disjoint union of the Θ(s) for s a sector-germ of I. Let θ ∈ Θ, q such that
θ ∈ Θ(q) and ǫ ∈ {−, +}. One says that θ is of sign ǫ if q is of sign ǫ. For ǫ ∈ {−, +}, one
denotes by Θǫ the set of elements of Θ of sign ǫ.
Theorem 5.7. Let q1 , q2 be two sector germs of I of the same sign. For i ∈ {1, 2}, let
θi ∈ Θ(qi ). Then there exists k ∈ R∗+ such that dθ1 ≤ kdθ2 . In particular for all θ ∈ Θ, the
topology of (I, dθ ) only depends on the sign of θ.

18

Proof. Let A be an apartment containing q1 and q2 , and d ∈ N (A). Let s0 = q1 , . . . , sn = q2
be a gallery joining q1 and q2 . For all i ∈ J0, nK, one sets ui = θsi (d). By an induction using
Lemma 5.6, there exists a ∈ R∗+ such that du0 ≤ adun . As every norms are equivalent on A,
there exists b, c ∈ R∗+ such that dθ1 ≤ bdu0 and dun ≤ cdθ2 , which concludes the proof of the
theorem.
We thus obtain (at most) two topologies on I: the topology T+ obtained by taking a
positive θ ∈ Θ and the topology T− obtained by taking a negative θ ∈ Θ. We will see that
when I is not a building, these topologies are different.
Corollary 5.8. Let A be an apartment of I and θ ∈ Θ. Then the topology of A induced by
the topology of (I, dθ ) is the affine topology on A.
Proof. By Theorem 5.7, one can suppose θ ∈ Θ(q) where q is a sector germ of A of the same
sign as θ. Then Lemma 5.1 concludes the proof.
Corollary 5.9. Let q be a sector-germ. Let ρ be a retraction with center q, A = ρ(I) and
d ∈ N (A). Then:
(i) for each θ ∈ Θ of the sign of q, ρ : (I, dθ ) → (A, d) is Lipschitz continuous
(ii) if B is an apartment and d′ ∈ N (B), ρ|B : (B, d′) → (A, d) is Lipschitz continuous.
Proof. Let θ ∈ Θ(q). Then by Lemma 4.9, ρ : (I, dθ ) → (A, dθ ) is Lipschitz continuous and
one concludes the proof of (i) with Theorem 5.7 and Lemma 5.1.
Let q′ be a sector-germ of B of the same sign as q and θ′ ∈ Θ(q′ ). Then ρ : (I, dθ′ ) → (A, d)
is Lipschitz continuous by (i). Thus ρ|B : (B, dθ′ ) → (A, d) is Lipschitz continuous and one
concludes with Lemma 5.1.
Corollary 5.10. Let A, B be two apartments of I. Then A ∩ B is a closed subset of A (seen
as an affine space).
Proof. By Lemma 3.12, A and B are closed for T+ (or T− ). Th A ∩ B is closed for T+ ,
therefore it is closed for the topology induced by T+ on A, and Corollary 5.8 completes the
proof.
Remark 5.11. Suppose that I is not a building. Then by Subsection 3.3, for all θ ∈ Θ,
(I, dθ ) is not complete.
Let q be a sector-germ of I and (Qn ) be an increasing sequence of sectors with germ q.
q
One says that (Qn ) is converging if there exists a retraction onto an apartment ρ : I → ρ(I)
such that (ρ(xn ))Sconverges, where xn is the base point of Qn for all n ∈ N and we call limit
of (Qn ) the set n∈N Qn . One can show that the fact that I is not complete implies that
there exists a converging sequence of direction q whose limit is not a sector of I, which is
impossible in a building. To prove this one can associate to each Cauchy sequence (xn ) a
sequence (zn ) such that d(zn , xn ) → 0 and such that zn + q ⊂ zn+1 + q for all n ∈ N. Then
we show that (zn ) converges in (I, dθ ) if an only if the limit of (zn + q) is a sector of I.

5.3

Study of the action of G

In this subsection, we show that each g ∈ G, g : I → I is Lipschitz continuous for the
distances we constructed. We begin by treating the case where g stabilizes a sector-germ.
After this, we treat the case where g stabilizes an apartment and then we conclude.
19

Lemma 5.12. Let g ∈ G and suppose that g stabilizes some sector-germ q. Let θ ∈ Θ(q).
Then g : (I, dθ ) → (I, dθ ) is an isometry.
Proof. Let us prove that g is 1-Lipschitz continuous. Let A be an apartment containing q
q
and φ : A → g.A. Let τ making the following diagram commute:
(A, dθ )❑

g

/ (g.A, dθ )
❑❑❑
❑❑φ❑
τ
❑❑❑
% 

.

(g.A, dθ )
Then τ is an automorphism of g.A stabilizing q and thus τ is a translation. As g.A ⊃ q, τ is
|g.A
an isometry of (g.A, dθ ). By Lemma 4.9, φ is an isometry. Thus g|A is an isometry and it
is in particular 1-Lipschitz continuous. By Lemma 5.5, g is 1-Lipschitz continuous. As g −1
stabilizes q, it is also 1-Lipschitz continuous, which shows the lemma.
Lemma 5.13. Let A be an apartment of I. We fix a structure of vectorial space on A. Let
|A
Q be a sector of I with base point 0. Let g ∈ G stabilizing A. Let w = g|A and w
~ be the
linear part of w. Then for all x ∈ I and u ∈ Q, g.(x + u) = g.x + w(u).
~
q

g.q

Proof. Let ρq : I → A and ρg.q : I → A. Let us show that ρg.q = g.ρq .g −1. Let x ∈ I
B∩A
and B be an apartment containing x and q. Let φ : B → A. One has ρq (x) = φ(x). As
g.q
g.φ.g −1 : g.B → A, ρg.q (g.x) = g.φ(g −1.g.x) = g.ρq (x). Therefore ρg.q = g.ρq .g −1 . Let x ∈ I
and u ∈ Q. One has
ρg.q (g(x + u)) = g.ρq (x + u) = w.(ρq (x) + u) = w(ρq (x)) + w(u).
~
We also have
ρg.q (g.x + w(u))
~
= ρg.q (g.x) + w(u)
~
= g.ρq (x) + w(u)
~
= ρg.q (g(x + u)).
Then B ∋ x + u and g.B ∋ g.(x + u), g.x + w
~ (u). As ρg.q|g.B is an isomorphism, g.(x + u) =
g.x + w(u).
~
Theorem 5.14. Let g ∈ G and θ ∈ Θ. Then g : (I, dθ ) → (I, dθ ) is Lipschitz continuous.
Proof. Let q be such that θ ∈ Θ(q). Let A be an apartment containing q and g.q. Let
g.q
φ : g.A → A and h ∈ G inducing φ on g.A. Then g = h−1 ◦ f , where f = hg. As h−1
is Lipschitz continuous by Lemma 5.12 (and by Theorem 5.7), it suffices to show that f is
Lipschitz continuous. One has f (A) = A. One fixes a vectorial structure on A. Let x, y ∈ I
|A
and w
~ be the linear part of f|A .
One writes θ = (dB )B∈A(q) . Let θ′ = θf.q (dA ). For x ∈ A, one sets |x| = dA (x, 0).
Let k ∈ R∗+ be such that w
~ : (A, | |) → (A, | |) is k-Lipschitz continuous. Let x, y ∈ I.
Let (u, v) ∈ TQ (x, y) such that |u| + |v| = dθ (x, y).
By Lemma 5.13, (w(u),
~
+
~
w(v))
~
∈ Tw.Q
~ (f (x), f (y)). Therefore, dθ ′ (f (x), f (y)) ≤ |w(u)|
|w(v)|
~
≤ kdθ (x, y). Therefore f : (I, dθ ) → (I, dθ′ ) is k-Lipschitz continuous and one concludes with Theorem 5.7.

20

5.4

Case of a building

In this subsection we suppose that I is a building. We show that the distances we constructed
and the usuals distance are equivalent.
If d is a W -invariant euclidean norm on A, one denotes by dI the extension of d S
to I in the
usual manner (see [Bro89] VI.3 for example). For x ∈ I, one denotes by st(x) = C∈A(x) C,
where A(x) is the set of alcoves that contains x. Then by a Lemma in VI.3 of [Bro89], st(x)
contains x in its interior.
Proposition 5.15. Let d be a W -invariant euclidean norm on A and θ ∈ Θ. Then there
exist k, l ∈ R∗+ such that kdI ≤ dθ ≤ ldI .
Proof. Let q be such that θ ∈ Θ(q). By Theorem 5.7, one can suppose that q ⊂ A and
that θ = θq (d). Let k, l ∈ R∗+ such that kd|A2 ≤ dθ|A2 ≤ ld|A2 . Let us first show that
Id : (I, dθ ) → (I, dI ) is k1 -Lipschitz continuous. Let A be an apartment containing q. Let
q
x, y ∈ A. Then dθ (x, y) = dθ (ρ(x), ρ(y)) ≤ k1 dI (ρ(x), ρ(y)) = k1 dI (x, y), where ρ : I → A.
Thus Id : (I, dθ ) → (I, dI ) is k1 -Lipschitz continuous by Lemma 5.5.
Let x, y ∈ I. As [x, y] is compact and thanks to the Lemma recalled before the proposition,
there exists n ∈ N∗ and x0 = x, x1 , . . . , xn = y ∈ [x, y] such that xi+1 ∈ st(xi ) for all
i ∈ J0, n − 1K. In particular, for all i ∈ J0, n − 1K, there exists an apartment Ai containing
xi , xi+1 and q. One has
dθ (x, y) ≤

n−1
X

dθ (xi , xi+1 ) =

i=0

n−1
X

dθ (ρ(xi ), ρ(xi+1 ))

i=0
n−1
X

≤l

dI (ρ(xi ), ρ(xi+1 )) = l

i=0

n−1
X

dI (xi , xi+1 ) = ldI (x, y),

i=0

which proves the proposition.

6
6.1

Combined distances
Comparison of positive and negative topologies

In this subsection, we show that T+ and T− are different when I is not a building. Let
I0 = G.0 be the set of vertices of type 0. For this we prove that I0 is composed of limit
points when I is not a building and then we apply finiteness results of [Héb16].
Proposition 6.1. Let θ ∈ Θ. Then I0 is discrete in (I, dθ ) if and only if I is a building.
Proof. Suppose that I is a building. By Proposition 5.15, we can replace dθ by a usual
distance on I. By Lemma 3.14 one has I0 ∩ A = Y , which is a lattice of A. Let η > 0 such
that for all x, x′ ∈ Y , d(x, x′ ) > η implies x = x′ . Let x, x′ ∈ I0 such that d(x, x′ ) < η.
Let A be an apartment of I containing x and x′ and g ∈ G such that g.A = A. Then
d(g.x, g.x′) < η and thus x = x′ .
Suppose now that I is not a building and thus that W v is infinite. By Theorem 5.7, one
can suppose that θ ∈ Θ(q), where q is a sector-germ of A. Let ǫ > 0. Let us show that
there exists x ∈ I0 such that dθ (x, 0) < 2ǫ and x 6= x′ . Let M0 be a true wall direction of
A such that for all consecutive walls M1 and M2 of direction M0 , dθ (M, M ′ ) < ǫ (such a
direction exists by Proposition 2.3 2). Let M be a wall such that dθ (0, M) < ǫ and such that
21

0∈
/ D, where D is the half apartment of A delimited by M and containing q. By the proof
A∩A
of Proposition 3.11, there exists an apartment A such that A ∩ A = D. Let φ : A → A and
x = φ(0). Let y ∈ M such that dθ (0, y) < ǫ. Then by Lemma 4.9, dθ (x, y) = dθ (0, y) and
thus d(x, 0) < 2ǫ. As x ∈
/ A, x 6= 0 and we get the proposition.
Remark 6.2. In fact, by Theorem 5.14, we have shown that when I is not a building, each
point of I0 is a limit point.
If B is an apartment and (xn ) ∈ B N , one says that (xn ) converges towards ∞, if for some
(for each) norm | | on B, |xn | → +∞.
Proposition 6.3. Suppose that I is not a building. Let ǫ ∈ {−, +} and δ = −ǫ. Let q
(resp. s) be a sector-germ of I of sign ǫ (resp. δ). Let θ ∈ Θ(q). We equip I with dθ . Let
ρ be a retraction centered at s and (xn ) ∈ I0N be an injective and converging sequence. Then
ρ(xn ) → ∞ in ρ(I). In particular ρ is not continuous.
Proof. By Theorem 5.7, one can suppose that there exists an apartment A containing q and
s such that q and s are opposite in A. Maybe composing ρ by an isomorphism of apartments
q
fixing s, one can suppose that ρ(I) = A. Let ρq : I → A. Let x = lim xn and y = ρq (x). Then
by Corollary 5.9, ρq (xn ) → y. Let YA = I0 ∩ A. Then YA is a lattice of A by Lemma 3.14.
As ρq (xn ) ∈ I0 ∩ A for all n ∈ N, ρq (xn ) = y for n large enough.
We also have ρ(xn ) ∈ YA for all n ∈ N. By Theorem 5.6 of [Héb16], for all a ∈ YA ,
−1
ρ ({a}) ∩ ρ−1
q ({y}) is finite. This concludes the proof of this lemma.
Corollary 6.4. If I is not a building, T+ and T− are different.
Remark 6.5. Proposition 6.3 shows that if θ, θ′ ∈ Θ have opposite signs, then each open
subset of (I, dθ ) containing a point of I0 is not bounded for dθ′ .

6.2

Combined distances

In this section we define and study combined distances.
Let Θc = Θ+ × Θ− . Let θ = (θ+ , θ− ) ∈ Θ2 . One sets dθ = dθ+ + dθ− .
Remark 6.6. Let θ ∈ Θc . Let f : I → I. Then if for some θ+ ∈ Θ+ and θ− ∈ Θ− , f :
(I, dθ+ ) → (I, dθ+ ) and f : (I, dθ− ) → (I, dθ− ) are Lipschitz continuous (resp. continuous),
then f : (I, dθ ) → (I, dθ ) is Lipschitz continuous (resp. continuous).
Let A be an apartment of I and f : I → A. Suppose that for some θ′ ∈ Θ, f : (I, dθ′ ) →
(A, dθ′ ) is Lipschitz continuous (resp. continuous). Then f : (I, dθ ) → (A, dθ ) is Lipschitz
continuous (resp. continuous).
Theorem 6.7. Let θ ∈ Θc . We equip I with dθ . Then:
(i) For all θ′ ∈ Θc , dθ and dθ′ are equivalent.
(ii) For all g ∈ G, g is Lipschitz-continuous.
(iii) All retraction of I centered at a sector-germ is Lipschitz continuous.
(iv) The topology induced on each apartment is the affine topology.
(v) There exists δ > 0 such that for all x, x′ ∈ I0 such that x 6= x′ , d(x, x′ ) ≥ δ.

22

Proof. The assertions (i) to (iv) are consequences of Remark 6.6, Theorem 5.7, Corollary 5.9,
Theorem 5.14 and Corollary 5.8. Let us prove (v). Let k ∈ R∗+ such that ρ+∞ and ρ−∞ are
k-Lipschitz continuous. Let α > 0 such that for all x, x′ ∈ Y , x 6= x′ implies d(x, x′ ) ≥ α.
Let x, x′ ∈ I0 such that x 6= x′ . By Corollary 4.4 of [Héb16], ρǫ (x) 6= ρǫ (x′ ) for some
ǫ ∈ {−∞, +∞} and thus d(x, x′ ) ≥ αk .
Lemma 6.8. Let θ ∈ Θc and a ∈ I. Let A be an apartment of I containing a. We fix
q+
an origin in A. Let Q+ and Q− be opposite sector-germs of A based at 0 and ρ+ : I → A,
q−
ρ− : I → A, where q− , q+ are the germs of Q− and Q+ . Let d be a distance induced by a norm
on A. Then there exists k ∈ R∗+ such that for all x ∈ I, dθ (a, x) ≤ k(d(a, ρ− (x))+d(a, ρ+ (x)).
Proof. One writes θ = (θ+ , θ− ). By Theorem 6.7 (i) and Lemma 5.1, one can suppose that
θ− = θq− (d), θ+ = θq+ (d) and d = dθ|A2 .
Let ν ∈ Q. Let T+ = Tν : I → A and T− = T−ν : I → A. By Corollary 4.2 and Remark
4.3 of [Héb16], T+ (x), T− (x) ≤ d(h(ρ− (x), h(ρ+ (x))) with h : A → R a linear function.
Therefore, there exists l ∈ R∗+ such that T+ (x), T− (x) ≤ ld(ρ− (x), ρ+ (x)) for all x ∈ I.
Let i ∈ {−, +} and x ∈ I. Then
di (x, ρ+ (x)) ≤ di (x, x + T (x)ν) + di (ρ+ (x) + Tν (x)ν, ρ+ (x)) ≤ 2ld(0, ν)di (ρ− (x), ρ+ (x)),
by definition of Ti .
One has,
d(a, x) = d+ (a, x) + d− (a, x) ≤ d− (a, ρ− (x)) + d− (ρ− (x), x) + d+ (a, ρ+ (x)) + d+ (ρ+ (x), x)
and thus d(a, x) ≤ (2ld(0, ν) + 1)(d(a, ρ− (x)) + d(a, ρ+ (x)).
Corollary 6.9. Let θ ∈ Θc and let us equip I with d = dθ . Then if X ⊂ I the following
assertions are equivalent:
(i) X is bounded
(ii) for all retraction ρ centered at a sector of I, ρ(X) is bounded
(iii) there exist two opposite sectors q+ and q− such that if ρq− and ρq+ are retractions
centered a q− and q+ , ρq− (X) and ρq+ (X) are bounded.
Moreover each bounded subset of I0 is finite.
Proof. By Theorem 6.7, (i) implies (ii), and it is clear that (ii) implies (iii). The implication
(iii) implies (i) is a consequence of Lemma 6.8. The last assertion is a consequence of (iii)
and of Theorem 5.6 of [Héb16].
−1
Corollary 6.10. The ρ−1
+ (U) ∩ ρ− (U) such that U is an open set of an apartment A and
ρ− and ρ− are retraction onto A centered at opposed sectors of A form a basis of Tc . In
particular Tc is the initial topology with respect to retractions centered at sector-germs.

Proof. This is a consequence of Lemma 6.8.

23

7

Contractibility of I

In this section we prove the contractibility of I for T+ , T− and Tc . By Theorem 5.7, by
symmetry of the role of the Tits cone and of its opposite, it suffices to prove that there exists
θ ∈ Θ(+∞) (where +∞ = germ∞ (Cfv )) such that (I, dθ ) is contractible and Remark 6.6 will
conclude for the contractibility of (I, Tc ).
One choosesLa basis (ei )i∈J of A such that for some J ′ ⊂ J, (ei )i∈J\J ′ is a basis of
v
∗
∗
Ain and
i∈I R+ ei ⊕ Ain . Let (ei ) be the dual basis of (ei ). For x ∈ A, one sets
P Cf =
∗
|x| = i∈J |ei (x)|. Let dA be the distance on A induced by | | and θ = θ+∞ (dA ). Let d = dθ .
One has dA = dθ|A2 .
One uses the maps yν : I → A and Tν : I → R, for ν ∈ Cfv defined in Subsection 2.7.
Lemma 7.1. Let ν ∈ Cfv and η = mini∈I e∗i (ν). Then Tν is
yν : (I, d) → (A, d|A2 ) is η1 |ν| + 1-Lipschitz continuous.

1
-Lipschitz
η

continuous and

Proof. Let x, x′ ∈ I and (u, u′) ∈ T (x, x′ ) such that d(x, x′ ) = |u| + |u′|. Then x+ Tν (x)ν ∈ A
and thus x′ + u′ + Tν (x)ν = x + u + Tν (x)ν ∈ A. As η1 |u|ν − u ∈ Cfv , x′ + (Tν (x) + η1 |u|)ν ∈ A.
′

)
and we get the first part of lemma.
Consequently, Tν (x′ ) ≤ Tν (x) + d(x,x
η
One has
d(yν (x), yν (x′ )) = d(ρ+∞ (x) + Tν (x)ν, ρ+∞ (x′ ) + Tν (x′ )ν)
≤ d(ρ+∞ (x) + Tν (x)ν, ρ+∞ (x′ ) + Tν (x)ν) + d(ρ+∞ (x′ ) + Tν (x)ν, ρ+∞ (x′ ) + Tν (x′ )ν)
= d(ρ+∞ (x), ρ+∞ (x′ )) + d(Tν (x)ν, Tν (x′ )ν),

because d is invariant by translation on A.
By Lemma 4.9, d(ρ+∞ (x), ρ+∞ (x′ )) ≤ d(x, x′ ). One also has d(Tν (x)ν, Tν (x′ )ν) ≤ |Tν (x)−
Tν (x′ )||ν| ≤ η1 |ν|d(x, x′ ) and we can conclude.
Remark 7.2. One can also prove that the maps (ν, x) 7→ Tν (x) and (ν, x) 7→ yν (x) are
locally Lipschitz continuous.


 I × [0, 1] → I



1
1
(x,
t)
→
7
x
+
ν
if
< Tν (x)
v
Proposition 7.3. Let ν ∈ Cf . One defines φν :
1
−
t
1
−
t



1


≥ Tν (x)
(x, t) 7→ yν (x) if
1−t
(where we consider that 01 = +∞ > t for all t ∈ R). Then φν is a strong deformation retract
on A.
Proof. Let x ∈ A and t ∈ [0, 1]. Then Tν (x) = 0 and thus φν (x) = yν (x) = x. Let
x ∈ I. Then φν (x, 1) = yν (x) ∈ A. It remains to show that φν is continuous. Let (xn , tn ) ∈
1
<
(I ×[0, 1])N be a converging sequence and (x, t) = lim(xn , tn ). Suppose for example that 1−t
1
1
1
Tν (x) (the case 1−t = Tν (x) and 1−t > Tν (x) are analogous). Then by Lemma 7.1, 1−tn <
Tν (xn ) for n large enough and thus by continuity of addition (Lemma 4.5), φν ((xn , tn )) =
1
1
ν = φν (x, t). Therefore, φν is continuous, which concludes the proof.
ν → x+ 1−t
xn + 1−t
n
Corollary 7.4. The masure (I, d) is contractible.

I × [0, 1] → I





1
(x, t) 7→ φν (x, 2t) if t ≤
. Then ψν is a strong
Proof. Let ν ∈ Q. One sets ψν :
2



1


(x, t) 7→ 2(1 − t)yν (x) if t >
2
deformation retract on {0}, which proves the corollary.
24

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