NONARCHIMEDEAN FLAG DOMAINS AND SEMISTABILITY

Harm Voskuil^{1} School of Mathematics and Statistics University of Sydney N.S.W. 2006 AUSTRALIA december 1998

Introduction.

Let K be nonarchimedean local field. Then K is a finite extension of ℚ_{p} or 𝔽_{q}((t)). Let K^{∘} be the valuation ring of K. We consider a simply connected absolutely almost simple linear algebraic group G defined over K^{∘}. For a parabolic subgroup P ⊂ G the corresponding projective homogeneous space is denoted by X := G/P. The variety X cannot always be defined over K^{∘}. For convenience we will assume in the introduction that it is defined over K^{∘}.

Let L be an ample line bundle on X. Again we will assume in this introduction that the embedding in a projective space determined by L is defined over K^{∘}. For each maximal K-split torus S ⊂ G ⊗ K we take the S linearization of L ⊗ K obtained by restricting the unique G ⊗ K linearization of L ⊗ K to S. In this paper we will study the analytic subspaces Y^{s} and Y^{ss} of X ⊗ 𝕂. Here 𝕂 denotes the completion of the separable algebraic closure of K. The space Y^{s} consists of the points in X ⊗ 𝕂 that are stable for all maximal K-split tori S ⊂ G ⊗ K. Similarly Y^{ss} consists of the points that are semistable for all maximal K-split tori.

Let B denote the building of G(K). We construct a G(K)-equivariant map I , that associates to each point x ∈ Y^{ss} a convex subset I(x) ⊂ B. The subset I(x)

Typeset by AMS-TEX

1This work has been supported by the Australian Research Council.

is bounded if and only if x ∈ Y^{s}. The map I is defined by using semistability over K^{∘}.

For groups G that are split over K^{∘} the image of the map I is studied in some detail. The proofs involve some combinatorics and the commutator relations for additive subgroups. In particular we need the existence of polyhedra with certain integrality properties. The existence of such polyhedra is guaranteed only if the root system Φ of G is of type A_{ℓ}, B_{ℓ}, C_{ℓ}, D_{ℓ} or E_{6} (See §4). Using these one gets a partial description of I(x) in §7 for these groups. If the image I(x) solely depends on the ”absolute values of the coordinates” of x then this partial description of I(x) is in fact a complete description of I(x) for x ∈ Y^{s} (See §8).

It is hoped that using the map I one can understand the spaces Y^{ss} and Y^{s} better. In particular one can construct affinoid coverings of Y^{s} (See [V.2] for an example). In the case Y^{s} = Y^{ss} a pure affinoid covering has been constructed in [PV] and [V1]. If Y^{s} = Y^{ss} then the quotient Y^{s}/Γ is proper for a discrete co-compact subgroup Γ ⊂ G(K). If Y^{s} ≠ Y^{ss} then the quotient Y^{s}/Γ is still separated, but it is not proper anymore.

Our terminology differs a little bit from [PV] and [V1]. We call Y^{s} a nonarchimedean flag domain. In [V1] one calls Y^{s} a flag domain if and only if Y^{s}/Γ is proper. In [PV] such spaces were called symmetric spaces. However it seems better to preserve the name symmetric spaces for those spaces Y^{s} with automorphic forms for Γ on it and such that X = G/P with P maximal parabolic.

From the point of view of rigid analytic uniformization the space Y^{ss} is not very useful, since the group Γ does not act discretely on Y^{ss} if Y^{ss} ≠ Y^{s}. However from a moduli perspective it is quite interesting. (See [Ra], [To] and [RZ]).

§1. Ample Line Bundles

1.1. We describe the relative and absolute root system of G. The connection between dominant weights and ample line bundles is described. Using this one can find the field over which a nonarchimedean flag domain can be defined. We also recall some criteria with which one can determine whether a point is (semi-)stable.

1.2. The absolute root system. Let G be a simply connected absolutely almost simple linear algebraic group defined over K^{∘}. Let K_{s} denote the separable closure of K and K_{s}^{∘} its ring of integers.

Let S ⊂ T ⊂ G be such that S ⊗ K is a maximal K-split torus and T a maximal torus defined over K^{∘}. The torus T and the group G both split over K_{s}. Let X (T) be the character group of T. Let Φ be the absolute root system of G. We choose a simple basis Δ of Φ.

The Weyl group of Φ is denoted by W. The group W acts on X (T) ⊗ ℝ. After choosing a W -invariant inner product on X (T)⊗ℝ, the Weyl group W is generated by the reflections in the hyperplanes orthogonal to the simple roots α ∈ Δ. Let N_{T} and Z_{T} be the normaliser and centraliser of the torus T in G. Then W ≅ N_{T} /Z_{T} (K_{s}).

Let α_{i}, i = 1, …, ℓ be the simple roots in Δ. Here ℓ = dim_{K_{s}}(T ⊗ K_{s}) is the absolute rank of G. Let w_{i} ∈ W denote the reflection in the hyperplane orthogonal to α_{i}.

The simple basis Δ determines a Borel subgroup B ⊂ G. Let Φ^{+} denote the set of positive roots for Δ. To each character α ∈ Φ belongs a T -stable additive subgroup U_{α} ⊂ G on which T acts with character α. One has B =< T, U_{α}|α ∈ Φ^{+} >.

Let I ⊂ {1, …, ℓ} be a non-empty subset. The subgroup W_{I} ⊂ W is the group generated by the reflections w_{i}, i ∉ I. The parabolic subgroup P_{I} ⊃ B is the group P_{I} := BW_{I}B. Any parabolic subgroup of G is conjugated to exactly one of the groups P_{I}.

1.3. The relative root system. We fix an ordering on X (S) and choose a compatible ordering on X (T). Let Δ_{0} denote the set of simple roots of G with respect to T that vanish on S. The relative root system Φ_{K} of G consists of the roots of X (S). The relative Weyl group W_{K} acts on Φ_{K}. One has W_{K} ≅ N_{S}/Z_{S}(K). Here N_{S} and Z_{S} are the normaliser and centraliser of S in G.

The Galois group H := Gal(K_{s}/K) acts on X (T), since T is split over K_{s}. For any element h ∈ H the image h(Δ) of Δ is again a simple basis of Φ. Hence there exists a unique element w ∈ W such that wh(Δ) = Δ. We put h^{∗} := w ∘ h for this element w ∈ W. We call this action of H on X (T) the twisted action. Let H^{∗} := {h^{∗}|h ∈ H}. Then H^{∗} is a finite group acting on X (T) that preserves the simple basis Δ.

1.4. Ample line bundles. Every ample line bundle L on X = G/P is very ample. So L determines an embedding X ↪ ℙ(V). Here ℙ(V) denotes the projectivisation of a G-module V. The module V is irreducible if Char(K) = 0, whereas if Char(K) > 0 this might not be the case. (See [J]). The G-module V is uniquely determined by its highest weight λ. We will denote this G-module by V_{λ}.

Let (−, −) be the W -invariant inner product on X (T) ⊗ ℝ. The fundamental weights ω_{i} of our root system Φ are determined by 2(ω_{i}, α_{j})/(α_{j}, α_{j}) = δ_{ij}. The ample line bundles L on X = G/P_{I} correspond to the modules V_{λ} with highest weight λ = ∑_{i∈I} n_{i}ω_{i} with n_{i} > 0 for all i ∈ I.

Let Λ_{r} denote the root lattice, i.e. the sublattice of X (T) generated by the roots α ∈ Φ. In [T.2] theorems 3.3 and 7.2 the irreducible representations of G that are defined over K are determined. From this one gets the linear representations of G. Furthermore these representations are defined over K^{∘}.

For each λ ∈ Λ_{r} one has a representation ρ_{λ} of G into GL(⊕_{σ∈H^{∗}}V_{σ(λ)}) that is defined over K^{∘}. (If the weight λ is not contained in Λ_{r} one might need a skew field D defined over K to define ρ_{λ}). For our purposes the representations ρ_{λ}, λ ∈ Λ_{r} suffice.

Note that ρ_{λ} is irreducible if Char(K) = 0, whereas if Char(K) = p > 0 this might not be the case. 1.5. Fields of definition. For each weight λ = ∑_{i∈I} n_{i}ω_{i} ∈ Λ_{r} we have a representation of G defined over K^{∘}. Let v_{λ} ∈ ⊕_{σ∈H^{∗}}V_{σ(λ)} be a vector contained in V_{λ}⊕ < 0 >, whose component in V_{λ} is a highest weight vector. The image X_{λ} of the orbit G · v_{λ} in ℙ(⊕_{σ∈H^{∗}}V_{σ(λ)}) is isomorphic to G/P_{I}. We have a variety X^{†} ⊂ ℙ(⊕_{σ∈H^{∗}}V_{σ(λ)}) defined over K^{∘}, whose connected components are X_{σ(λ)}, σ ∈ H^{∗}. All the connected components of X^{†} are isomorphic. The very amply line bundle L^{†} associated with this embedding restricts on each connected component X_{σ(λ)} to the line bundle L belonging to the weight σ(λ).

For each X = G/P_{I} defined over K_{s}^{∘} and ample line bundle L on X corresponding to some weight λ ∈ Λ_{r}, one has a variety X^{†} and an ample line bundle L^{†}, both defined over K^{∘}, such that one connected component of X^{†} is isomorphic to X and L^{†}|_{X} ≅ L.

One easily sees that X^{†}⊗L^{∘} has its connected components defined over L^{∘}. Here L is a splitting field for G. Moreover also L^{†}|_{X⊗L^{∘}} is defined over L^{∘}. One can find a minimal field in L such that this is the case. This field will be called the field of definition of (X, L). 1.6. (Semi)-stable points. Let L be an ample line bundle on X corresponding to a weight λ, then L^{⊗n} corresponds to the weight n · λ. For every weight λ one can find an integer n > 0 such that n · λ ∈ Λ_{r}. Moreover L and L^{⊗n} determine the same sets of (semi-)stable points. So the assumption that λ ∈ Λ_{r} poses no real restriction.

Let X^{s}(T, L) and X^{ss}(T, L) denote the varieties consisting of the stable and the semistable points, respectively, in X for the action of T w.r.t. L. The T linearization is chosen to be the restriction of the unique G linearization of L (or better L^{†}). These sets can be determined using the criteria given in [MFK].

Let L give an embedding of X into ℙ(V) for some G-module V = V_{λ}. We have a decomposition V = ⊕V^{β}, β ∈ X (T) into eigenspaces V^{β} on which T acts with character β. Each V^{β} has a free K^{∘}-basis e_{1}^{β}, ⋯, e_{m(β)}^{β}. These bases determine coordinates x_{β,i} for every x ∈ X ⊂ ℙ(V). We may assume that x is normalised, i.e. |x_{β,i}| ≤ 1, ∀β, i and max|x_{β,i}| = 1. Then one defines the polyhedron μ(x) ⊂ X (T) ⊗ ℝ as the convex hull of {β|∃i |x_{β,i}| = 1}. One has for x ∈ X:

x ∈ X^{s}(T, L) ⟺ 0 ∈ int μ(x)
x ∈ X^{ss}(T, L) ⟺ 0 ∈ μ(x)

Part a) of the proposition below is proved in [GS] and [FH]. Part b) was proved in [PV] theorem 1.1:

1.7. Proposition. Let L be an ample line bundle on X = G/P corresponding to the weight λ. Let T ⊂ G be a maximal torus. Then one has:

a) For any point x ∈ X the vertices of μ(x) form a subset of the set {w(λ)|w ∈ W }. The edges of μ(x) are parallel to the roots α ∈ Φ. b) X^{s}(T, L) = X^{ss}(T, L) if and only if λ is not contained in a hyperplane

(through 0) spanned by roots.

1.8. Let r : X (T) ⟶ X (S) denote the restriction map. The sets of (semi- ) stable points X^{s}(S, L) and X^{ss}(S, L) can be described using the polyhedron r(μ(x)) ⊂ X (S) ⊗ ℝ. Below we recall some results and their proof from [V.1].

1.9. Proposition. Let L be an ample line bundle on X = G/P, corresponding to the weight λ ∈ Λ_{r} and let S ⊂ T ⊂ G as before. Then one has:

a) For any point x ∈ X the vertices of r(μ(x)) form a subset of {r(w(λ))|w ∈ W }. The edges of r(μ(x)) are parallel to roots α ∈ Φ_{K}.

b) X^{s}(S, L) = X^{ss}(S, L) if for all w ∈ W the weight r(w(λ)) is not contained in a hyperplane (through 0) spanned by roots α ∈ Φ_{K}.

c) X^{s}(S, L) ≠ X^{ss}(S, L) if λ is contained in a hyperplane V ⊂ X (T) ⊗ ℝ spanned by roots α ∈ Φ and moreover there exists an element w ∈ W such that r(w(V)) is contained in a hyperplane (spanned by roots β ∈ Φ_{K}).

Proof. Since r maps Φ into Φ_{K} ∪ {0}, part (a) of the proposition is clear. Furthermore (b) follows from (a).

To prove (c) one constructs for λ in a hyperplane V spanned by roots α ∈ Φ a point x ∈ X such that μ(x) = V ∩ conv ({w(λ)|w ∈ W }). This has been done in [PV] 1.4. Then one chooses an element w ∈ W such that r(w(V)) is contained in a hyperplane in X (S) ⊗ ℝ. Then 0 ∈ r(w(μ(x)) = r(μ(w(x))) ⊂ r(w(V)) and the point w(x) is in X^{ss}(S, L) − X^{s}(S, L). This proves (c).

1.10. For the convenience of the reader we recall the following theorem from [V1] without proof:

1.11 Theorem. There exists an ample line bundle L on X = G/P such that X^{ss}(S, L) = X^{s}(S, L) if and only if one of the following holds:

(1) P = B and G is any group.

(2) P = P_{J}, J = {1, …, ℓ − 1} and G is a non-split group with absolute root system Φ of type C_{ℓ}. Here α_{ℓ} is the unique long root in the simple basis Δ.

(3) P = P_{I} and G = SL_{s+1}(D) with g.c.d.(i ∈ I, s + 1) = 1. Here D is a skew field (of dimension d^{2}) defined over K (we allow D = K) and the P_{{i}} are such that G/P_{{i}} ≅ Gr(i, (s + 1) · d).

§2. Equivariant maps from the (semi-) stable points to the apartment

2.1. We fix a group G, a projective homogeneous variety X = G/P and an ample line bundle L on X corresponding to some weight λ ∈ Λ_{r}. Let S ⊂ G be a maximal K^{∘}-split torus and A ⊂ B the apartment corresponding to S.

The completion of the separable closure of K will be denoted by 𝕂. In this section we construct an S(𝕂)-equivariant map I_{A} : X^{ss}(S, L)(𝕂) ⟶ { convex subsets of A }. The map I_{A} will be studied in some detail.

First we recall the construction of the apartment A belonging to S(𝕂). 2.2. The apartment. Let X_{∗}(S) be the space of one-parameter subgroups (1-ps), i.e. the dual of the character space X (S). Let V := X_{∗}(S) ⊗ ℝ. The perfect pairing between characters and 1-ps will be denoted by < −, − >. We extend it to a pairing of V and X (S) ⊗ ℝ.

One defines a map ν : S ⟶ V by < χ, ν(s) >= −v(χ(s)) for all χ ∈ X (S). Here v is the additive valuation of K, normalized such that v(π) = 1 for a generator π of the maximal ideal. We regard V as an affine space. The torus S acts on V by translation, as follows: s · z = z + ν(s) for z ∈ V.

To G and S belongs an affine root system Φ_{aff}. The affine roots are functions α + n : V ⟶ ℝ with α ∈ Φ_{K} and n ∈ Γ_{α}. Here (α + n)(z) = α(z) + n for z ∈ V, where α(z) :=< α, z >. Furthermore Γ_{α} is the discrete group ℤ or \frac{1}{2}ℤ, depending on the particular affine root system Φ_{aff} and the root α.

The affine root system gives a simplicial decomposition of V. The maximal simplices or chambers are the closures of the connected components of {z ∈ V |∀(β ∈ Φ_{aff}) β(z) ≠ 0}. The affine space V endowed with this simplicial decomposition is the apartment A for the torus S ⊗ K. The map ν defined above extends uniquely to a map from S(𝕂) to A. Hence we have a well defined action of S(𝕂) on A. 2.3. Analytification of the sets of (semi-) stable points. To each algebraic variety corresponds a rigid analytic variety that has the same set of closed points.

(See [BGR] or [FP]). We denote the analytifications of X^{s}(S, L)⊗𝕂 and X^{ss}(S, L)⊗ 𝕂 by Y_{A}^{s} and Y_{A}^{ss}, respectively. For notational convenience we have dropped the line bundle from the notation (though these spaces do depend on L). The subscript corresponds to the apartment A belonging to S(K).

To each point z ∈ A we want to associate an analytic subspace Y_{z,A}^{s} (resp. Y_{z,A}^{ss}) of Y_{A}^{s} (resp. Y_{A}^{ss}). One defines Y_{0,A}^{s} (resp. Y_{0,A}^{ss}) as being the completion of X^{s}(S, L) ⊗ 𝕂^{∘} (resp. X^{ss}(S, L) ⊗ 𝕂^{∘}) along the closed fiber. Here 𝕂^{∘} is the ring of integers of 𝕂. One has by construction:

Y_{0,A}^{s}(𝕂) = X^{s}(S, L)(𝕂^{∘})
Y_{0,A}^{ss}(𝕂) = X^{ss}(S, L)(𝕂^{∘}) .

If z ∈ A is a point such that z = s · 0 for some s ∈ S(𝕂) then one defines:

Y_{z,A}^{s} := s · Y_{0,A}^{s} = {s · x|x ∈ Y_{0,A}^{s}}
Y_{z,A}^{ss} := s · Y_{0,A}^{ss} = {s · x|x ∈ Y_{0,A}^{ss}} .

Since X^{ss}(S, L)/S is a projective algebraic variety, the map ψ^{ss} : S(𝕂)×Y_{0,A}^{ss} ⟶ Y_{A}^{ss}, defined by ψ^{ss}(s, x) = s · x, is surjective. The map ψ^{s} : S(𝕂) × Y_{0,A}^{s} ⟶ Y_{A}^{s}, (s, x) ⟶ s · x, is surjective if and only if Y_{A}^{s} = Y_{A}^{ss}.

Note that both Y_{z,A}^{s}, and Y_{z,A}^{ss} are quasi-compact, i.e. they are the union of finitely many affinoid subspaces.

2.4. Definition. For x ∈ Y_{A}^{ss} we define the interval of S-semistability I_{A}(x) as follows:

I_{A}(x) : = {z ∈ A|x ∈ Y_{z,A}^{ss}}
= {s^{−1} · 0 ∈ A|s · x ∈ Y_{0,A}^{ss}}

Note that I_{A}(s · x) = s · I_{A}(x) for all s ∈ S(𝕂) and x ∈ Y_{A}^{ss}.

2.5. Proposition.

a) I_{A}(x) ⊂ A is convex.

b) I_{A}(x) is bounded if and only if x ∈ Y_{A}^{s}.

c) I_{A}(x) = {z} if and only if x ∈ Y_{z,A}^{s}.

Proof. Since I_{A}(s · x) = s · I_{A}(x) for s ∈ S(𝕂), we may assume that 0 ∈ I_{A}(x), i.e. x ∈ Y_{0,A}^{ss}. We assume that x ∈ Y_{0,A}^{ss} is normalised. Let the coordinates x_{β,i} be as in 1.6 and let r : X (T) ⟶ X (S) be the restriction map.

One has : s · x ∈ Y_{0,A}^{ss} ⟺ |r(β)(s) · x_{β,i}| ≤ 1 ∀β, i. Hence s ∈ S(𝕂) is such that s · x ∈ Y_{0,A}^{ss} satisfies:

− v(r(β)(s)) ≤ v(x_{β,i}) ∀ β, i s.t. x_{β,i} ≠ 0 .
⟺ − v(r(β)(s)) ≤ min {v(x_{β,i})| x_{β,i} ≠ 0} ∀β s.t. ∃i with x_{β,i} ≠ 0 .

From this one concludes that ν(s) has to be in the intersection of a finite number of affine half spaces. The same is true of I_{A}(x) = {s^{−1} ·0|s·x ∈ Y_{0,A}^{ss}}. Hence I_{A}(x) is convex. This proves a).

The set I_{A}(x) is bounded if and only if the convex hull of {r(β)|∃i x_{β,i} ≠ 0} contains 0 in its interior. By definition this is equivalent to x ∈ Y_{A}^{s}. This proves b).

Now c) follows from b). Indeed, I_{A}(x) = {0} if and only if {r(β)|∃i |x_{β,i}| = 1} has the same convex hull as {r(β)|∃i x_{β,i} ≠ 0} and has 0 in the interior of the convex hull. Hence I_{A}(x) = {0} if and only if x ∈ Y_{0,A}^{s}.

2.6. Proposition. The interval of semistability I_{A}(x) for x ∈ Y_{A}^{ss} is bounded by hyperplanes parallel to walls. In particular there exist n_{α}(x) ∈ ℝ ∪ {∞}, α ∈ Φ_{K} such that:

I_{A}(x) = {z ∈ A|∀(α ∈ Φ_{K}) α(z) ≤ n_{α}(x)} .

Proof. Without loss of generality we may assume that x ∈ Y_{0,A}^{ss}. Furthermore we may assume that 0 is not contained in a face of the polyhedron I_{A}(x). We normalise x such that max {|x_{β,i}|| β ∈ X (T)} = 1. Let V_{1}(x, A) ⊂ X (S) ⊗ ℝ be the linear subspace spanned by the weights r(β), β ∈ X (T) such that for some i one has |x_{β,i}| = 1. From proposition 1.9 if follows that either V_{1}(x, A) = 0 or V_{1}(x, A) is spanned by roots α ∈ Φ_{K}.

If V_{1}(x, A) = X (S) ⊗ ℝ we are done. Indeed, then x ∈ Y_{0,A}^{s} and I_{A}(x) = {0}. Hence we can take n_{α}(x) = 0 for all α ∈ Φ_{K}.

So let us assume that V_{1}(x, A) ≠ X (S)⊗ℝ. Let S_{1} ⊂ S be the subtorus on which all characters χ ∈ V_{1}(x, A) ∩ X (S) vanish. Clearly if for s ∈ S, one has s · x ∈ Y_{0,A}^{ss}, then s ∈ S_{1}. Hence I_{A}(z) ⊂ {z ∈ A|α(z) = 0 ∀(α ∈ V_{1}(x, A) ∩ Φ_{K})}.

Let s ∈ S_{1} be such that s^{−1}·0 is in the interior of a codimension one face of I_{A}(x). Let F ⊂ X (S)⊗ℝ be the linear subspace by the r(β) such that some |x_{β,i}(s·x)| = 1. Then by proposition 1.9 a) the space F is generated by roots α ∈ Φ_{K}. Moreover dim (F) = dim (V_{1}(x, A)) + 1. Hence one can find a root α ∈ F ∩ Φ that together with V_{1}(x, A) spans F. The face of I_{A}(x) containing s^{−1} · 0 is the intersection of I_{A}(x) with a hyperplane defined by α(z) = α(s^{−1} · 0). Furthermore I_{A}(x) is either contained in the half space {z ∈ A| α(z) ≤ α(s^{−1} · 0} or in the halfspace {z ∈ A| α(z) ≥ α(s^{−1} · 0}. Since we can do this for any codimension one face of I_{A}(x), the proposition follows. 2.7. Remark. Note that I_{A}(x) = A if and only if x ∈ Y_{A}^{ss} is a fixed point for the action of S(𝕂). Such a fixed point exists if and only if r(w(λ)) = 0 for some w ∈ W. In particular this never occurs if G is split.

If none of the characters r(w(λ)), w ∈ W vanishes on S, then I_{A}(x) ⊂ A has always some codimension > 0.

For later use we describe the codimension one faces of I_{A}(x) in the lemma below. 2.8. Lemma. Let G be a split group. Let V_{1}(x, A) ⊂ X (S) ⊗ ℝ be the linear subspace spanned by the set of characters that are constant on I_{A}(x). Then the codimension one faces of I_{A}(x) correspond to roots α ∈ Φ such that α ⊥̸ V_{1}(x, A), that is the faces are given by α(z) = c_{α}(x). Proof. Note that we have on X (S) ⊗ ℝ a W -invariant inner product that is unique up to a scalar.

Again we may assume that 0 is in I_{A}(x), but not on a face of I_{A}(x). Furthermore we may assume that x ∈ Y_{0,A}^{ss} is normalised. Then V_{1}(x, A) =< μ|∃i |x_{μ,i}| = 1 >. Let s ∈ S be such that s^{−1} · 0 is in a codimension one face H of I_{A}(x). We may choose s in such a way that s^{−1} · 0 is not contained in a codimension two face of I_{A}(x).

Now s · x ∈ Y_{0,A}^{ss}. Let F :=< μ|∃i |x_{μ,i}(s · x)| = 1 > as in the proof of proposition 2.6. Since both F and V_{1}(x, A) are spanned by roots and dim (F) = dim V_{1}(x, A) + 1, we can find a root α in F − V_{1}(x, A). Clearly the face H of I_{A}(x) is given by α(z) = c_{α}(x), for a suitable constant c_{α}(x) ∈ ℝ.

So we only have to show that we can choose α ⊥̸ V_{1}(x, A). This follows from the fact that the w ∈ W with |x_{w(λ)}(sx)| = 1 are the vertices of a convex polyhedron with edges parallel to roots. Moreover if λ is joined to w(λ) by an edge parallel to α then w = w_{α}, the reflection associated to α. In particular α ⊥̸ λ, w_{α}(λ).

§3 Equivariant maps to the building

3.1. Introduction. Let Y^{s} := ⋂_{g∈G(K)} g(Y_{A}^{s}) and Y^{ss} := ⋂_{g∈G(K)} g(Y_{A}^{ss}).

We will construct a G(K)-equivariant map I from Y^{ss} to the set of convex subsets of B. Here B is the Bruhat-Tits building of G(K).

First we recall the definition of the building B. To each point z ∈ B we associate some analytic subspaces Y_{z}^{s}, Y_{z}^{ss} ⊂ X(𝕂). One proves that Y_{z}^{s} ⊂ Y^{s} and Y_{z}^{ss} ⊂ Y^{ss}. Then one uses the spaces Y_{z}^{ss} to define I(x). 3.2. The Building. All maximal K-split tori S ⊂ G(K) are conjugated. To each maximal K-split torus gSg^{−1}, g ∈ G(K), belongs an apartment g(A). Each simplex σ ∈ A is stabilized by a parahoric subgroup P_{σ} ⊂ G(K). The simplex g(σ), g ∈ G(K), is stabilized by P_{g(σ)} := gP_{σ}g^{−1}.

The building B is defined as B := ⋃_{g∈G(K)} g(A)/ ∼. The equivalence relation ∼ is given by σ_{1} ∼ σ_{2} if and only if P_{σ_{1}} = P_{σ_{2}}. Here σ_{1}, σ_{2} are simplices in the building B. 3.3. The intervals of G(K)-semistability. Let z′ ∈ A′. We can find an element g ∈ g(K) such that A′ = g(A). We put z := g^{−1}(z′) ∈ A. Then Y_{z′,A′}^{s} := g(Y_{z,A}^{s}) and Y_{z′,A′}^{ss} := g(Y_{z,A}^{ss}) .

We associate the following analytical subspaces of X(𝕂) to a point z ∈ B:

Y_{z}^{s} := ⋂_{A′∋z} Y_{z,A′}^{s} = ⋂_{g∈P_{z}} g(Y_{z,A′}^{s}) .
Y_{z}^{ss} := ⋂_{A′∋z} Y_{z,A′}^{ss} = ⋂_{g∈P_{z}} g(Y_{z,A′}^{ss}) .

Here P_{z} ⊂ G(K) denotes the stabilizer of z ∈ B. Note that P_{z} = P_{σ}, where σ ∈ B is the smallest simplex containing the point z.

For x ∈ Y^{ss} one defines the interval of G(K)-semistability I(x) as follows:

I(x) := {z ∈ B|x ∈ Y_{z}^{ss}} .

Since Y_{z}^{ss} = ⋂_{A′∈z} Y_{z,A′}^{ss} one has:

I(x) = {z ∈ B|∀(A′ ∈ z) z ∈ I_{A′}(x)} .

Note that I_{gA}(x) = g(I_{A}(g^{−1}(x)) for g ∈ G(K). Clearly one has by construction that I(g(x)) = g(I(x)) for g ∈ G(K).

3.4. Definition. Let f_{1}, …, f_{m} be generators of the set of S-invariants Γ(X, L^{⊗d})^{S}. Here d is chosen in such a way that the homogeneous S-invariants generate ⊕_{n>0}Γ(X, L^{⊗dn})^{S} as a K^{∘}-algebra.

Let A_{1} = g_{1}(A) and A_{2} = g_{2}(A) be two different apartments and g_{1}, g_{2} ∈ G(K). As in [PV] section 3.6 we define a function r_{A_{1},A_{2}} : Y_{A_{2}}^{ss} ⟶ ℝ as follows:

r_{A_{1},A_{2}}(x) = r_{g_{1}(A),g_{2}(A)}(x) : max_{1≤i≤m}{|g_{1}^{∗}f_{i}(x)|}/ max_{1≤i≤m}{|g_{2}^{∗}f_{i}(x)|} .

This function has the following obvious properties:

a) r_{A_{1},A_{2}}(x) = r_{A_{2},A_{1}}(x)^{−1}, if x ∈ Y_{A_{1}}^{ss} ∩ Y_{A_{2}}^{ss}.

b) r_{g(A_{1}),g(A_{2})}(g(x)) = r_{A_{1},A_{2}}(x) for g ∈ G(K) and x ∈ Y_{A_{2}}^{ss}.

Below we will study the map r_{A_{1},A_{2}} in more detail. We follow [PV] closely here. 3.5. Lemma. Let x ∈ Y_{z,A}^{ss} and assume that z ∈ A ∩ A′. Then r_{A′,A}(x) ≤ 1. Proof. Since z is contained in some simplex σ ∈ A ∩ A′, one can apply the proof of [PV] 3.6 property c).

3.6. Definition. For x ∈ X(𝕂) we define the function r̃: X(𝕂) ⟶ ℝ by:

r̃(x) = \{ \begin{matrix}0 & if x ∉ Y_{A}^{ss} \\ inf \{r_{gA,A}(x)|g ∈ G(K)\} & if x ∈ Y_{A}^{ss} \end{matrix}

3.7. Proposition. Let x ∈ Y_{A}^{ss}. There exists an apartment A′ ⊂ B such that r_{A′,A}(x) = r̃(x)

Proof. See [PV] 3.6, proof of property d).

3.8. Proposition. Let z ∈ A′ ⊂ B and x ∈ Y_{z,A′}^{ss}. Then x ∈ Y_{z}^{ss} if and only if r_{A′,A}(x) = r̃(x) > 0.

Proof. We first prove that if r̃(x) = 0 then x ∉ Y_{z}^{ss}. If r̃(x) = 0 then by proposition 3.7 we can find an apartment A_{1} ⊂ B such that r_{A_{1},A}(x) = 0. Hence x ∉ Y_{A_{1}}^{ss}. Now we can find z_{1} ∈ A_{1} and an apartment A_{2} such that x ∈ Y_{z_{1},A_{2}}^{ss}. Since A_{1} = hA_{2} for some h ∈ P_{z_{1}} we can construct h̃ ∈ P_{z} such that r_{h̃A_{2},A_{2}}(x) < ϵ. In particular we may assume that 0 < r_{h̃A_{2},A}(x) < r_{A′,A}(x). Now x ∈ Y_{z_{2},h̃A_{2}}^{ss} for some z_{2} ∈ h̃A_{2}.

Let A′′ be such that z_{2}, z ∈ A′′ Using lemma 3.5 one finds that r_{A′′ ,A}(x) < r_{A′,A}(x). It follows in particular that x ∉ Y_{z,A′′}^{ss} . Hence x ∉ Y_{z}^{ss}.

Now let us assume that r̃(x) > 0. Let A_{1} ⊂ B be an apartment such that r_{A_{1},A}(x) = r̃(x). Suppose x ∈ Y_{z_{1},A_{1}}^{ss}. Let A_{2} be an apartment containing both z and z_{1}. Using lemma 3.5 one finds r_{A_{2},A}(x) = r̃(x) > 0. Again applying 3.5 gives us x ∈ Y_{z}^{ss} ⟺ r_{A′,A}(x) = r̃(x) > 0.

3.9. Corollary. Let z ∈ B. Then Y_{z}^{ss} ⊂ Y^{ss}.

Proof. If x ∈ Y_{z}^{ss} then r̃(x) > 0 by proposition 3.8. Hence for all A′ ⊂ B one has x ∈ Y_{A′}^{ss}. Thus x ∈ Y^{ss}.

3.10. Theorem. Y^{ss} = ⋃_{z∈B} Y_{z}^{ss}.

Proof. In the corollary above we have already shown that Y_{z}^{ss} ⊂ Y^{ss}. Therefore it is sufficient to prove that Y^{ss} ⊂ ⋃_{z∈B} Y_{z}^{ss}.

If x ∈ Y^{ss} then there exists an apartment A′ ⊂ B such that r_{A′,A}(x) = r̃(x). Hence x ∈ Y_{z,A′}^{ss} for some z ∈ A′. Using proposition 3.8 one concludes that x ∈ Y_{z}^{ss}. This proves the theorem.

3.11. Proposition. Let x ∈ X(𝕂). Then x ∈ Y^{ss} if and only if r̃(x) > 0.

Proof. Clear from proposition 3.8 and theorem 3.10.

3.12. Proposition. Let x ∈ Y^{ss}. Then I(x) = ⋃{I_{A′}(x)|r_{A′,A}(x) = r̃(x)}

Proof. This follows immediately from the definition of I(x) and propositions 3.8 and 3.11.

3.13. Corollary. Let x ∈ Y^{ss}. Then I(x) is convex.

Proof. Let z_{1}, z_{2} ∈ I(x). If A′ ⊂ B is an apartment containing both z_{1} and z_{2} then r_{A′,A}(x) = r̃(x). Hence I_{A′}(x) ⊂ I(x). Now z_{1}, z_{2} ∈ I_{A′}(x) and I_{A′}(x) is convex. From this the corollary follows.

3.14. Theorem. Let x ∈ Y^{ss}. Then x ∈ Y^{s} if and only if I(x) is bounded.

Proof. We will first show that if I(x) is not bounded then x ∉ Y^{s}. So let us assume I(x) is not bounded. Let z ∈ I(x). We can find a sequence of points z_{i} ∈ I(x) such that the distance dist (z, z_{i}) ⟶ ∞ if i ⟶ ∞. Let A_{i} ⊂ B be an apartment containing both z and z_{i}.

Then I_{A_{i}}(x) ⊂ I(x) and z, z_{i} ∈ I_{A_{i}}(x). There exists h_{i} ∈ P_{z} such that A_{i} = h_{i}A. Here we assumed that z ∈ A. Since P_{z} is compact, a subsequence of the sequence h_{i} converges to an element h ∈ P_{z}. Clearly I_{hA}(x) ⊂ I(x) and I_{hA}(x) is not bounded. Hence x ∉ Y_{hA}^{s}. So x ∉ Y^{s}. This proves the “only if” part of the theorem.

Next we will show that if x ∈ Y^{ss} − Y^{s} then the interval I(x) is not bounded. Let x ∈ Y^{ss} − Y^{s}. There exists an apartment Ã such that x ∈ Y ss ; Ã − Y s ; Ã. In particular IÃ(x) is not bounded. Let z ∈ I(x). Then there exists an apartment A′ such that z ∈ A′ and A′ ∩ IÃ(x) is not bounded (see [B] Ch. VI §8). Now I_{A′}(x) is not bounded by lemma 3.15 below. Furthermore I_{A′}(x) ⊂ I(x), since z ∈ A′. Hence I(x) is not bounded.

3.15. Lemma. Let x be contained in Y_{A}^{ss} ∩ Y_{A′}^{ss}. Suppose that both I_{A}(x) and I_{A}(x) ∩ A′ are not bounded. Then I_{A′}(x) is not bounded.

Proof. Since I_{A}(x) and A′ are convex, I_{A}(x) ∩ A′ is convex. Let z ∈ I_{A}(x) ∩ A′. There exists a 1 − ps. μ : 𝕂 ⟶ S(𝕂) such that μ(s) · z ∈ I_{A}(x) ∩ A′ for all s ∈ 𝕂 such that |s| ≤ 1.

Let g ∈ P_{z} be such that A′ = g · A. We may assume that g stabilises A ∩ A′. Then g ∈ P_{μ(s)·z} for s such that |s| ≤ 1. Let σ_{0} ∈ A be a chamber containing z. Let σ_{i} := μ(π^{i}) · σ_{0}, i ≥ 0. Then g ∈ P_{σ_{i}}, i ≥ 0.

Since g ∈ P_{σ_{i}}, one has g = t · ∏_{α∈Φ_{K}} u_{α}(f_{α}), with |f_{α}| ≤ n_{α,i}. Here we have taken

a fixed order for the roots α ∈ Φ_{K}, and t ∈ S(𝕂^{∘}). One has P_{σ_{i}} = μ(π^{i})P_{σ_{0}}μ(π^{i})^{−1} and μ(π^{i})u_{α}(f_{α})μ(π^{i})^{−1} = u_{α}(α(μ(π^{i})) · f_{α}). Therefore n_{α,i} ⟶ 0 if i ⟶ ∞, for any root α such that α(μ) > 0. Hence we have g = t · ∏_{α∈Φ_{K}, <α,μ>≤0} u_{α}(f_{α}).

By assumption we have x ∈ Y_{μ(s)·z,A}^{ss} if |s| ≤ 1. Therefore μ(s)^{−1} · x ∈ Y_{z,A}^{ss} if |s| ≤ 1. This implies that x_{χ,j} = 0 for all χ such that r(χ)(μ) > 0.

One has u_{α}(f_{α})^{∗}x_{χ,j} = x_{χ,j} + ∑_{ν} a_{ν,i}(f_{α})x_{ν,i}, where r(ν) = r(χ) − nα for some

n ∈ ℤ_{≥0}.

From this one concludes that g^{∗}x_{χ,i} = 0 if r(χ)(μ) > 0 for our point x. Hence g · μ(s) · g^{−1} · y ∈ I_{A′}(x) if |s| ≤ 1 and y ∈ I_{A′}(x). In particular I_{A′}(x) is not bounded. (Note that I_{A′}(x) intersected with A might be empty!).

3.16. Corollary. A point x ∈ Y^{ss} is contained in Y_{z}^{s} if and only if I(x) = {z}. In particular Y_{z}^{s} ⊂ Y^{s}.

Proof. A point x ∈ Y^{ss} is in Y_{z}^{s} if and only if for all apartments A containing z, one has I_{A}(x) = {z}. This happens only when I(x) = {z}. This proves the first statement.

Let x ∈ Y_{z}^{s}. Then I(x) = {z} and hence is bounded. By theorem 3.14 the point x is in Y^{s}. Therefore Y_{z}^{s} ⊂ Y^{s}.

3.17. Remark. Let F ⊂ B be a bounded subset. Let cc(F) be the circumcenter of F. The circumcenter of F is the center of the closed ball with minimal radius that contains F. It is well defined and unique (See [B] Ch. VI §4). This gives a G(K)-equivariant map ψ : Y^{s} ⟶ B, where ψ(x) = cc(I(x)). This map cannot be extended to all of Y^{ss} if Y^{ss} ≠ Y^{s}. This clarifies somewhat property 3.12 in [Ra].

§4. Ideal polyhedra for split groups

4.1. Introduction. In this section we associate to a linear subspace V ⊂ X (S)⊗ℝ that is spanned by roots, certain polyhedra Δ(V, J, R, z) with radius R, center z and a ”parameter set” J (See 4.2). Our main interest is in such polyhedra that have certain integrality properties. These polyhedra will be called ideal polyhedra. For the root systems Φ = A_{ℓ}, …, D_{ℓ} and E_{6} we prove the existence of such ideal polyhedra. For the other irreducible root systems the existence of ideal polyhedra depends on the choice of the linear space V.

In the next section these polyhedra will be used to give a stratification of I_{A}(x).

4.2. Definition. Let the group G be split. Let S = T be a maximal K^{∘} split torus in G. We fix a W -invariant inner product (−, −) on X (S) ⊗ ℝ. The inner product is unique up to a scalar. We fix this scalar such that the long roots α ∈ Φ satisfy (α, α) = 2 if Φ ≠ C_{ℓ}, ℓ ≥ 3, G_{2}, (α, α) = 4 if Φ = C_{ℓ}, ℓ ≥ 3 and (α, α) = 3 if Φ = G_{2}. Using the inner product we identify X (S) ⊗ ℝ with the apartment A = X (S)_{∗} ⊗ ℝ belonging to S.

Let Φ be the root system belonging to G. Let Φ^{∘} ⊂ Φ be equal to Φ if Φ ≠ C_{ℓ}, ℓ ≥ 3 and consist of the short roots if Φ is of type C_{ℓ}, ℓ ≥ 3. Let V ⊂ X (S) ⊗ ℝ be a linear subspace spanned by roots. We will always assume that V ≠< 0 >.

We define the space V^{c} as being the linear space spanned by the roots that are orthogonal to all roots that are orthogonal to all roots in V, that is V^{c} :=<< V^{⊥} ∩ Φ >^{⊥} ∩Φ >. If V^{⊥} ∩ Φ = ∅, then V^{c} := X (S) ⊗ ℝ. Since we assume that V ≠< 0 >, < V^{⊥} ∩ Φ >^{⊥} ∩Φ = ∅ does not occur.

For each root α ∈ V^{c} ∩ Φ^{∘}, such that α ∉ V, we fix an integer p_{α} ∈ ℤ_{>0}. Let J := {(p_{α}, α) |α ∈ Φ^{∘} ∩ V^{c}, α ∉ V }. If V^{c} = V then J := ∅. The root polyhedron Δ(V, J, R, z) ⊂ A with parameter set J, radius R and center z is defined as the union of all the points y ∈ A that satisfy the following conditions:

1) ∀(χ ∈ V) (χ, y) = (χ, z)

2) ∀(α ∈ Φ^{∘}, α ∉ V^{c}, α ∉ V^{⊥}) (α, y − z) ≤ R
3) ∀(α ∈ Φ^{∘}, α ∉ V, α ∈ V^{c}) (α, y − z) ≤ p_{α} · R

4.3. Remark. Different parameter sets J, J′ can (and quite often do) determine the same root polyhedron. In many cases the polyhedron, obtained by omitting condition 3) of the definition, is already bounded. In these cases the polyhedron is the same for any choice of the parameters p_{α} such that p_{α} ≥ m_{α} for suitably choosen integers m_{α} > 0.

The lemma below explains our choice of Φ^{∘} for the root system of type C_{ℓ}, ℓ ≥ 3.

4.4. Lemma. Let Φ ⊂ ℝ^{ℓ} be a root system of type C_{ℓ}, ℓ ≥ 3. Let V ⊂ ℝ^{ℓ} be a linear subspace spanned by roots. Let α ∈ Φ be a root such that α ⊥̸ V and α ∉ V. Then there exists a short root β ∈ Φ , such that α ≡ β mod V.

Proof. Let Φ consist of the roots ±2e_{i}, ±e_{i} ± e_{j}, i ≠ j, 1 ≤ j ≤ ℓ. Here the e_{i}, i = 1, …, ℓ form an orthonormal basis of ℝ^{ℓ}.

Let α ∈ Φ be such that α ⊥̸ V and α ∉ V. If α is short then we can take β = α and we are done. So let us assume that α is long. Then there exists a root γ ∈ V such that (α, γ) > 0. Clearly γ has to be a short root. Then β := α − γ is again a short root and satisfies the statement of the lemma.

4.5. Definition. Let V ⊂ X (S) ⊗ ℝ be a linear subspace spanned by roots. We put c_{α,V,J} := max{(α, y)/R| y ∈ Δ(V, J, R, 0)}. We are mainly interested in root polyhedra that have the property that c_{α,V,J} ≥ 1 for all α ∈ Φ^{∘} such that α ∉ V, V^{⊥}. A root polyhedron that has this property will be called a decent root polyhedron . A decent root polyhedron Δ(V, J, R, z) is called minimal if every decent polyhedron Δ(V, J′, R, z) contained in Δ(V, J, R, z) is equal to Δ(V, J, R, z) itself. A minimal decent root polyhedron will be called an ideal polyhedron.

4.6. Proposition. Let Φ ≠ F_{4} or G_{2}. Let α ∈ Φ^{∘}, α ∉ V^{c} and α ∉ V^{⊥}. Then c_{α,V,J} = 1.

Proof. From the definition of Δ(V, J, R, 0) it is immediate that c_{α,V,J} ≤ 1 for α as above. So to proof the proposition it is sufficient to show that there exists for a root α ∉ V^{c}, V^{⊥} a point y ∈ Δ(V, J, R, 0) such that (α, y) = R.

Since α ∉ V^{c}, there exists a root β ∈ V^{⊥} such that (α, β) ≠ 0. Let y := (R/(α, β)) · β. Then (α, y) = R. Since (γ, y) ≤ R for any root γ ∈ Φ^{∘}, γ ≠ ±β, it follows that y ∈ Δ(V, J, R, 0).

4.7. Remark. If α ∈ Φ^{∘}, α ∉ V and α ∈ V^{c}, the value of c_{α,V,J} can be different for different roots α. Consider the following example.

Let Φ = D_{7}, consisting of the roots ±e_{i} ± e_{j}, i ≠ j, 1 ≤ i, j ≤ 7. Here the e_{i}, i = 1, …, 7 form an orthonormal basis of ℝ^{7}. Let α_{i} := e_{i} − e_{i+1}, i = 1, …, 6 and α_{7} := e_{6} + e_{7}. The α_{i}, 1 ≤ i ≤ 7 form a simple basis of Φ. Let V be the hyperplane generated by the α_{i}, i ≠ 3. The parameter set J is determined by taking p_{α} = 2 for all roots α ∉ V. This determines a root polyhedron Δ(V, J, R, z).

The linear space V^{⊥} is spanned by the fundamental weight ω_{3} = e_{1} + e_{2} + e_{3}. One has V^{c} = ℝ^{7} ≠ V. Let α ∈ V^{c} − V be a root. Then α = Σ_{i}n_{i} · α_{i}, with n_{3} ≠ 0. In fact n_{3} is either ±1 or ±2. Therefore |(ω_{3}, α)| is either 1 or 2. Hence Δ(V, J, R, 0) = {t · ω_{3}| − R ≤ t ≤ R}. It follows that c_{α,V,J} = 2 if n_{3} = ±2 and that c_{α,V,J} = 1 if n_{3} = ±1.

4.8. Lemma. Let α ∈ Φ, α ∉ V, α ∉ V^{⊥}. We moreover assume that p_{α} = 2 for all roots α ∈ Φ^{∘} ∩ V^{c}, α ∉ V. Then the following statements hold:

i) If Φ = B_{ℓ} and α is short then c_{α,V,J} = 1.

ii) If Φ = C_{ℓ}, ℓ > 2 and α is long then c_{α,V,J} = 2.

Proof. Since the proof of both statements is rather similar, we only give a proof of statement (i). Let Φ = {±e_{i} ± e_{j}, ±e_{i}| i ≠ j, 1 ≤ i, j ≤ ℓ} be the root system of type B_{ℓ}, ℓ > 1. Here the e_{i} form an orthonormal basis of ℝ^{ℓ}. We may assume that α = e_{1}. Since α ∉ V and α ∉ V^{⊥}, there exists a root β ∈ V such that (α, β) ≠ 0. We may assume that β = e_{1} − e_{2}.

From the fact that e_{1} ∉ V, one concludes that the irreducible component Φ^{†} of Φ ∩ V that contains β is a root system of type A_{m} for some m ≥ 1. Moreover, this is the only irreducible component of the root system Φ ∩ V that is not orthogonal to α = e_{1}. Without loss of generality, we may assume that Φ^{†} consists of the roots ±(e_{i} − e_{j}) with i ≠ j and 1 ≤ i, j ≤ m + 1 ≤ ℓ.

If m = 1, then e_{1} + e_{2} ∈ V^{⊥} and α ∉ V^{c}. One verifies that y := R · (e_{1} + e_{2}) ∈ Δ(V, J, R, 0). Moreover, (α, y) = R. Hence c_{α,V,J} = 1 in this case.

Now let us consider the case where m > 1. Then α ∈ V^{c}. Let y := R · (e_{1} + e_{2} + ⋯ + e_{m+1}). One verifies that y ∈ Δ(V, J, R, 0). One has (α, y) = R. We leave it now to the reader to verify that c_{α,V,J} = 1.

4.9. Proposition. Let V ⊂ X (S) ⊗ ℝ be a linear subspace spanned by roots. We assume that V^{⊥}∩Φ = ∅ and that V^{⊥} ≠< 0 >. Then there exists an ideal polyhedron Δ(V, J, R, z) with radius R and center z.

Proof. It is sufficient to prove the proposition for the case where R = 1 and z = 0. We construct an ideal polyhedron.

Choose a simple basis Δ of Φ, consisting of the simple roots α_{1}, …, α_{ℓ} , such that V is spanned by simple roots. Let W′ ⊂ W be the subgroup consisting of the elements w ∈ W such that w(Δ∩V) = Δ∩V. Let W_{V} be the Weyl group of Φ∩V. Then W′ · W_{V} is the normaliser N_{W} (W_{V}) of W_{V} in W (See [Ho] Corr. 3).

Let O_{1}, …, O_{m} be the orbits of N_{W} (W_{V}) on roots α ∉ Φ. Let Δ_{i,n} be the polyhedron defined by (β, z) ≤ n for all β ∈ O_{i} and (α, z) = 0 for all α ∈ V. Now let us define n_{i}, i = 1, …, m inductively as being the smallest integer in ℤ_{>0} such that max{ (α, z) | z ∈ Δ_{1,n_{1}} ∩ ⋯ Δ_{i,n_{i}}} ≥ 1 for all roots α ∈ Φ^{∘} that are not in V. Now we define J by taking p_{α} = n_{i} for α ∈ O_{i}. Then Δ(V, J, 1, 0) = Δ_{1,n_{1}} ∩ ⋯ Δ_{m,n_{m}}.

It is clear from the construction that p_{α} = c_{α,V,J} ≥ 1 for any root α ∈ Φ−Φ∩V. Hence Δ(V, J, 1, 0) is a decent root polyhedron.

If we replace any p_{α} by a smaller number, then it is clear that the thus obtained root polyhedron is not decent anymore. From this one concludes that our polyhedron does not contain any smaller decent root polydra. Therefore Δ(V, J, 1, 0) is an ideal polyhedron.

4.10. Lemma. Let Φ be a root system of type A_{ℓ}. Let J be given by putting for all α ∈ Φ ∩ V^{c}, α ∉ V the parameter p_{α} equal to p_{α} = 1. Then for all roots α ∉ V, V^{⊥} one has that c_{α,V,J} = 1.

Proof. If ℓ = 1 there is nothing to prove. So let us consider the case where Φ is of type A_{ℓ}, ℓ > 1. We choose a simple basis α_{1}, …, α_{ℓ} of the root system Φ, such that the linear space V is spanned by simple roots. Let α_{i} be a simple root that is not contained in V. Then the fundamental weight ω_{i} is orthogonal to V. For any root α one has (α, ω_{i}) = 0, ±1. Indeed, this follows from the fact that α = Σn_{j} · α_{j} with n_{j} ∈ {0, ±1}. Therefore ±R · ω_{i} ∈ Δ(V, J, R, 0) for any i such that α_{i} ∉ V. From this it easily follows that for any root α that is not contained in V and V^{⊥} one has that c_{α,V,J} = 1.

4.11. Lemma. Let Φ be a root system of type B_{ℓ}, C_{ℓ} with ℓ ≥ 2 or D_{ℓ} with ℓ > 3. We put p_{α} = 2 for all α ∈ Φ^{∘} ∩ V^{c}, α ∉ V. Then for all roots α ∉ V, V^{⊥} one has that c_{α,V,J} ≥ 1.

Proof. Since the roots α ∉ V^{c}, V^{⊥} have been treated in proposition 4.6, we only have to consider roots α that are contained in V^{c} but not in V.

Let us first consider the root system D_{ℓ} with ℓ > 3. Let Φ consist of the roots ±e_{i} ± e_{j} with i ≠ j and 1 ≤ i, j, ≤ ℓ. Here the e_{i} form an orthonormal basis of ℝ^{ℓ}. First we look at the case where V^{⊥} ∩ Φ ≠ ∅. Then V^{c} ≠ V if and only if Φ ∩ V has at least one irreducible component of type A_{i} with i > 1. If i = 3 then we moreover have to assume that the component of type A_{3} is contained in a root system of type A_{4} in D_{ℓ}. Let us choose such a component of type A_{i}. We may assume that it consists of the roots ±(e_{n} − e_{m}) with 1 ≤ n < m ≤ i + 1. One verifies that the point y := R · (e_{1} + ⋯ e_{i+1}) is contained in the polyhedron Δ(V, J, R, 0). Hence for all roots α ∉ V, V^{⊥} that are not orthogonal to our component of type A_{i} one has (y, α) = ±R, ±2R. In particular c_{α,V,J} ≥ 1.

To each irreducible component Φ_{j} of Φ ∩ V that is of type A_{i_{j}}, we can associate a point y_{j} as above. If i_{j} = 3 then we make the same assumption as above. Let α ∈ V^{c} be a root. Then either α is not orthogonal to exactly one component of type A_{i_{j}}, i_{j} > 1, or to exactly two components of type A_{i_{j}}, i_{j} > 1, or to one component of type A_{i_{j}}, i_{j} > 1 and one component of type D_{i_{j}}, i_{j} > 1. (Here D_{2} = A_{1} × A_{1} and D_{3} = A_{3} are such that they are not contained in a system of type A_{4} inside Φ.) Then c_{α,V,J} is 2, 2 and 1, respectively.

Now let us consider the case where Φ = D_{ℓ} and V^{⊥} ∩ Φ = ∅. We choose a simple basis α_{1}, …, α_{ℓ} of Φ such that V is spanned by simple roots. Let α_{i} ∉ V and ω_{i} be the corresponding fundamental weight. For each root α ∈ Φ one has (α, ω_{i}) = 0, ±1, ±2. Hence y_{i} := R · ω_{i} is contained in Δ(V, J, R, 0). Since this is true for all roots α_{i} that are not contained in V, it follows that c_{α,V,J} ≥ 1 for all roots α ∉ V.

For the root systems of type B_{ℓ}, respectively C_{ℓ}, the short respectively long roots have been treated above. The proof for the remaining roots is similar to that for the root system D_{ℓ} and we will omit it.

4.12. Lemma. Let Φ ⊂ ℝ^{6} be a root system of type E_{6}. Let V ⊂ ℝ^{6} be a linear space spanned by roots such that V^{⊥} ∩ Φ ≠ ∅. For each root α ∈ Φ ∩ V^{c} − Φ ∩ V we take p_{α} = 1. Then c_{α,V,J} = 1.

Proof. We fix a simple basis α_{1}, …, α_{6} of Φ. The ordering of the roots is such that (α_{i}, α_{i+1}) = (α_{1}, α_{4}) = −1 for i = 2, …, 5. Since V^{⊥} ∩ Φ ≠ ∅ , we may assume that the highest root α_{0} is contained in V^{⊥}. We may moreover assume that V is spanned by simple roots. Then Φ ∩ V is a root system that is contained in the root system of type A_{5} that is generated by the roots α_{2}, …, α_{6}.

If Φ ∩ V is a root system of type A_{5}, then V^{⊥} =< α_{0} >. Therefore V = V^{c} and the proposition follows from proposition 4.6. Now let us assume that rank(Φ∩V) = 4. Then a simple basis of Φ ∩ V is obtained by ommitting a simple root α_{j} from the basis of the root system of type A_{5}. We will denote the corresponding space V by V_{j}. If j = 4, then V^{⊥} =< α_{0}, α_{1} >. Hence V_{4} = V_{4}^{c} and we are done.

Let us now assume that j ≠ 4. Let the highest root α_{0} in E_{6} be α_{0} = Σn_{i} · α_{i}. Define μ_{j} as being μ_{j} := n_{1} · ω_{j} − n_{j} · ω_{1} = 2 · ω_{j} − n_{j} · ω_{1}. Here the ω_{i} are the fundamental weights belonging to our simple basis. Note that μ_{j} ⊥ α_{0} and that ω_{1} = −α_{0}. We have n_{j} = 2 if j = 3, 5 and n_{j} = 1 if j = 2, 6. We will now show that 1/2 · μ_{j} is in Δ(V_{j}, J, 1, 0).

If α ∈ Φ^{+} and α ∉ V^{c}, V^{⊥}, then α = Σm_{i} · α_{i} with m_{1} = 1 and m_{j} ≤ n_{j}. Then (α, μ_{j}) = 2 · m_{j} − n_{j}. So (α, μ_{j}) ∈ {0, ±2} if n_{j} = 2 and (α, μ_{j}) ∈ {0, ±1} if n_{j} = 1. Now let α ∈ Φ^{+} be such that α ∈ V^{c} − V. Then α = Σm_{i} · α_{i} with m_{1} = 0, m_{j} = 1 and m_{i} = 0, 1 if i ≠ 1, j. From this it follows that (α, μ_{j}) = 2. Therefore 1/2 · μ_{j} ∈ Δ(V_{j}, J, 1, 0) . Now it is clear that c_{α,V_{j},J} = 1 for all roots α ∉ V.

Let us now consider the case where rank(Φ ∩ V) = 3. Then Φ ∩ V is of type A_{3}, A_{1} × A_{2} or of type A_{1} × A_{1} × A_{1}. If Φ ∩ V is of type A_{3} , then Φ ∩ V^{⊥} is of type A_{1} × A_{1}. In this case one has V = V^{c}, even though dim(V) + dim(V^{⊥}) = 5. So in this case we are done by using proposition 4.6.

If Φ ∩ V is of type A_{1} × A_{2}, then Φ ∩ V^{c} is of type A_{2} × A_{2}. We may choose V =< α_{3}, α_{5}, α_{6} >. Using the weight ω_{2} one now shows that c_{α,V,J} = 1.

If Φ ∩ V is of type A_{1} × A_{1} × A_{1}, then Φ ∩ V^{c} is of type A_{5}. One may choose V =< α_{2}, α_{4}, α_{6} >. Now one can use the weights ω_{3} − ω_{1} and ω_{5} − ω_{1} to show that c_{α,V,J} = 1 in this case.

If rank(Φ∩V = 2) then Φ∩V = A_{2} (and Φ∩V^{⊥} = A_{2} ×A_{2}) or Φ∩V = A_{1} ×A_{1} (and Φ∩V^{⊥} = A_{3}). In both cases V = V^{c}. If rank(Φ∩V) = 1, then Φ∩V^{⊥} = A_{5}. So again we have V = V^{c} and we are done. We have now treated all the cases. 4.13. Theorem. Let Φ ⊂ ℝ^{ℓ} be an irreducible root system. Let V ⊂ ℝ^{ℓ} be a linear space spanned by roots. We assume that V ≠< 0 >, ℝ^{ℓ}. If Φ is of type A_{ℓ}, B_{ℓ}, C_{ℓ}, D_{ℓ} or E_{6} then ideal polyhedra Δ(V, J, R, z) exist. If Φ ∩ V^{⊥} ≠ ∅, then these ideal polyhedra are unique. Proof. The existence of the ideal polyhedra in these cases has been proved in the lemmas above. So we only have to prove the uniqueness in the case where Φ∩V^{⊥} ≠ ∅.

If V = V^{c}, then the uniqueness is clear (See also lemma 4.6). So let us assume that V ≠ V^{c}. If there exist ideal polyhedra with c_{α,V,J} = 1 for all roots α ∈ Φ^{∘}, α ∉ V, V^{⊥}, then the uniqueness is obvious. This holds for Φ = A_{ℓ}, E_{6} by lemmas 4.10 and 4.12. Moreover it follows from the proof of lemma 4.11 that this also holds for Φ = D_{ℓ} if Φ ∩ V contains no component of type D_{i} (Here we follow the same conventions as in that proof).

In the other cases the ideal polyhedron is obtained by taking p_{α} = 2 for all roots α ∈ Φ^{∘}, α ∈ V^{c}, α ∉ V. One sees that in these cases the ideal polyhedron in fact is entirely determined by the properties 1) and 2) of definition 4.2. From this the uniqueness follows. 4.14. Remark. For the irreducible root systems not treated in the proposition above ideal polyhedra do not always exist. The existence depends on V.

Let W_{V} be the Weyl group of Φ ∩ V. The normaliser N_{W} (W_{V}) of W_{V} preserves the ideal polyhedra Δ(V, J, R, 0) for the root systems A_{ℓ}, …, D_{ℓ}, E_{6}. If Φ∩V^{⊥} = ∅ then it follows from the construction in proposition 4.9. If Φ∩V^{⊥} ≠ ∅ then it follows from the uniqueness.

4.15. Remark. In the paragraphs that follow we will only consider ideal polyhedra. In vieuw of the uniqueness of these polyhedra if V^{⊥} ∩ Φ ≠ ∅, we will drop the parameter set J from the notation. If V^{⊥} ∩ Φ = ∅, then we fix a parameter set (which has the same properties as those constructed in proposition 4.9). So from now on we will write Δ(V, R, z) and c_{α,V} instead of Δ(V, J, R, z) and c_{α,V,J}.

From now on we will restrict ourselves to the root systems Φ of type A_{ℓ}, B_{ℓ}, C_{ℓ}, D_{ℓ} and E_{6}. The proofs given are still valid for the root systems E_{7} and E_{8}, provided that Y^{ss} is such that for all the linear spaces V_{i}(x, A) (See def. 5.3) that occur, ideal polyhedra Δ(V_{i}(x, A), R, z) exist. For the root systems of type F_{4} and G_{2} it is not clear, wether our definition of ideal polyhedra is the correct one.

§5. A stratification of I_{A}(x) for split groups

5.1. Introduction. In this section we study I_{A}(x) in some detail for split groups G with root system Φ of type A_{ℓ}, …, D_{ℓ} or E_{6}. In particular we give a stratification I_{A}^{i}(x) of I_{A}(x). Here I_{A}^{i}(x) ⊂ I_{A}^{i−1}(x) is a certain convex subset if i > 1 and I_{A}^{1}(x) := I_{A}(x).

The stratification is defined by using suitable ideal polyhedra Δ(V, R, z) ⊂ I_{A}^{i}(x). Here V is spanned by the roots that are constant on I_{A}^{i}(x) and the radius R is taken maximal. These ideal polyhedra Δ(V, R, z) resemble I_{A}^{i}(x) itself, but are in general more symmetric. If z = 0 then the group N_{W} (W_{V})/W_{V} acts on the ideal polyhedron. As I_{A}^{i+1}(x) one takes the union of the centres z of all polyhedra Δ(V, R, z) with maximal radius R that are contained in I_{A}^{i}(x).

Using this stratification one is able to compare I_{A}(x) and I_{A′}(x) for apartments A and A′. This will be done in §7, where we describe I(x) in some detail.

5.2. Definition. Let F ⊂ A be a convex polyhedron. We admit polyhedra that are not bounded. Let V (F) ⊂ X (S) ⊗ ℝ be the linear subspace spanned by the characters χ that are constant on F. We assume that V (F) is spanned by roots.

The radius R(F) of F is defined as R(F) := max {R|Δ(V (F), R, z) ⊂ F }. If the maximum does not exist we say that R(F) = ∞.

If R(F) < ∞ then the set of centres of F is given by Cent(F) := {z|Δ(V (F), R(F), z) ⊂ F }. If R(F) = ∞, then the set Cent(F) is not defined.

The set Cent(F) is again a convex polyhedron, if it exists. Moreover the dimension of Cent(F) is less than that of F. If the faces of F are parallel to walls (i.e. if the faces are of the form {y ∈ F |(α, y) = c_{α}} for some c_{α} ∈ ℝ and α ∈ Φ) then V (Cent(F)) is again spanned by roots, if R(F) < ∞.

We will now apply this to I_{A}(x).

5.3. Definition. We again fix a projective homogeneous variety X for G and an ample line bundle L on it. Let x ∈ Y_{A}^{ss}. We construct a strictly increasing flag V_{•}(x, A) = (V_{1}(x, A) ⊂ ⋯ ⊂ V_{m(x,A)}(x, A)) of linear subspaces of X (S) ⊗ ℝ. The length m(x, A) ≥ 1 depends on x (and A). Associated to each V_{i}(x, A) is a polyhedron I_{A}^{i}(x) that is contained in I_{A}(x). Their definition will be done inductively.

First we define V_{1}(x, A) := V (I_{A}(x)) and I_{A}^{1}(x) := I_{A}(x). Let us now assume that V_{i}(x, A) and I_{A}^{i}(x) are defined. If V_{i}(x, A) = X (S)⊗ℝ or R(I_{A}^{i}(x)) = ∞, then we put m(x, A) := i and we are done.

So let us assume that V_{i}(x, A) ≠ X (S) ⊗ ℝ and that R(I_{A}^{i}(x)) < ∞. Then we define I_{A}^{i+1}(x) := Cent(I_{A}^{i}(x)) and V_{i+1}(x, A) := V (I_{A}^{i+1}(x)).

For i < m(x, A) the subset I_{A}^{i+1}(x) of I_{A}^{i}(x) is a polyhedron of smaller dimension. Therefore everything is well defined.

The linear space V_{m(x,A)}(x, A) is equal to the entire space X (S) ⊗ ℝ if and only if I_{A}(x) is bounded. This is the case if and only if x ∈ Y_{A}^{s}.

5.4. Proposition. Let x ∈ Y_{0,A}^{ss} be such that 0 is not contained in a face of I_{A}(x). We assume that x is normalised and that |x_{λ}| = 1. Let 1 ≤ i ≤ m(x, A). Then there exists in each irreducible component of the root system Φ ∩ V_{i}(x, A) a root α such that (α, λ) ≠ 0. In particular if i > 1 then each irreducible component of Φ ∩ V_{i}(x, A) contains at least one irreducible component of Φ ∩ V_{i−1}(x, A).

Proof. First we consider the case i = 1. Then V_{1}(x, A) is spanned by the weights w(λ), w ∈ W such that |x_{w(λ)}| = 1. Since λ ∈ V_{1}(x, A), the weights w(λ) that occur are such that the element w ∈ W is in fact an element of W′. Here W′ is the Weyl group of the root system Φ ∩ V_{1}(x, A). Therefore the weights w(λ), w ∈ W′ span V_{1}(x, A). From this it follows that λ is not orthogonal to any irreducible component of Φ ∩ V_{1}(x, A).

Next we consider the case where i > 1. The linear space V_{i}(x, A) is generated by V_{i−1}(x, A) and the roots α that correspond to certain faces of I_{A}(x). By lemma 2.8 these roots α are never othogonal to V_{1}(x, A). Using induction it follows that each irreducible component of Φ ∩ V_{i}(x, A) contains at least one irreducible component of Φ ∩ V_{i−1}(x, A). In particular each irreducible component contains at least one irreducible component of Φ ∩ V_{1}(x, A). Therefore λ is not orthogonal to any irreducible component of Φ ∩ V_{i}(x, A) for 1 ≤ i ≤ m(x, A).

5.5. Definition. We make the correspondence between faces I_{A}^{i}(x) and roots a little bit more precise. A codimension one face F of I_{A}^{i}(x) is said to correspond to a root α if there exists a constant c ∈ ℝ such that F = {z ∈ I_{A}^{i}(x)|(α, z) = c} and if for all y ∈ I_{A}^{i}(x) one has (α, y) ≥ c.

Now we are able to state:

5.6. Proposition. Let us fix for each face F of I_{A}^{i}(x) a root β_{F} ∈ Φ^{∘} that corresponds to the face F. We assume that R(I_{A}^{i}(x)) < ∞. Let Φ_{i}^{∘} consist of the roots β_{F} corresponding to faces F that have the property that every Δ(V_{i}(x, A), R(I_{A}^{i}(x)), z) ⊂ I_{A}^{i}(x) has a non-empty intersection with F.

Then the β_{F} ∈ Φ_{i}^{∘} together with V_{i}(x, A) generate V_{i+1}(x, A). Moreover there exist n_{β} > 0, β ∈ Φ_{i}^{∘} such that ∑_{β∈Φ_{i}^{∘}} n_{β} · β = 0 mod V_{i}(x, A).

Proof. Let Φ^{i} consist of the roots β that correspond to a face F of I_{A}^{i}(x). The root β ∈ Φ^{i} is constant on I_{A}^{i+1}(x) if and only if the face F of I_{A}^{i}(x) corresponding to β has the same distance to every center z of an ideal polyhedron Δ(V_{i}(x, A), R(I_{A}^{i}(x)), z) ⊂ I_{A}^{i}(x). This is clearly the case for the β ∈ Φ_{i}^{∘}. Hence V_{i+1}(x, A) contains Φ_{i}^{∘}.

If β is not in the span of Φ_{∘}^{i} and V_{i}(x, A), then one can find two different polyhedra Δ(V_{i+1}(x, A), R(I_{A}^{i}(x)), z_{j}), j = 1, 2, such that the corresponding centers z_{j}, j = 1, 2 have a different distance to the face corresponding to β. Hence V_{i+1}(x, A) is spanned by V_{i}(x, A) and Φ_{i}^{∘}.

Next we prove the second statement of the proposition. Suppose there do not exist n_{β} > 0 for the β ∈ Φ_{i}^{∘} such that ∑ n_{β} · β = 0 mod V_{i}(x, A). Let H ⊂ Φ_{i}^{∘} be the largest subset such that there do exist n_{β} > 0, β ∈ H such that ∑ n_{β} · β = 0 mod V_{i}(x, A). Clearly H is unique and by assumption we have that H ≠ Φ_{i}^{∘}. For each β ∈ Φ^{i}, β ∉< Φ_{i}^{∘}, V_{i}(x, A)) one can find an ideal polyhedron Δ(V_{i}(x, A), R(I_{A}^{i}(x)), z_{β}) ⊂ I_{A}^{i}(x) such that this polyhedron has a non-empty intersection with the face F corresponding to β. Let z be in the convex hull of the z_{β}. We may assume that z is not on a face of the polyhedron formed by the convex hull of the z_{β}. Then Δ := Δ(V_{i}(x, A), R(I_{A}^{i}(x)), z) ⊂ I_{A}^{i}(x). Moreover the only faces F that have non-empty intersection with Δ are the faces F corresponding to roots β_{F} ∈ Φ_{i}^{∘}.

Let s ∈ S(𝕂) be such that χ(s) = 1 for all χ ∈< V_{i}(x, A), H >. We choose s in such a way that |β(s)| > 1 for β ∈ Φ_{i}^{∘} − H and ϵ ≤ |β(s)| ≤ ϵ^{−1} for some 0 < ϵ < 1. By choosing ϵ close enough to 1, we can make sure that s · Δ ⊂ I_{A}(x), since no relation ∑ n_{β} · β = 0 mod < V_{i}(x, A), H >, n_{β} > 0, β ∈ Φ_{i}^{∘} − H exists. This cannot be. Therefore we must have H = Φ_{∘}^{i}. This proves the second statement of the proposition.

5.7. Definition. We define a function ρ_{A} : Y_{A}^{ss} ⟶ ℝ^{ℓ} that measures the volume of I_{A}(x) for x ∈ Y_{A}^{ss}. Here ℓ is the rank of G. Let ρ_{A}(x) = (ρ_{A}(x)_{i}|i = 1, …, ℓ) and R_{j} := R(I_{A}^{j}(x)), j = 1, …, m(x, A). Then we define:

ρ_{A}(x)_{i} := \{ \begin{matrix}1 & if 0 < i ≤ dim(V_{1}(x, A)). \\ exp (− ∑_{s=1}^{j} R_{s}) & if dim(V_{j}(x, A)) < i ≤ dim(V_{j+1}(x, A)) . \\ 0 & if dim(V_{m(x,A)}(x, A)) < i. \end{matrix}

Here exp denotes the exponent function.

On ℝ^{ℓ} we put the lexicographical ordering, i.e. a = (a_{i}) < b = (b_{i}) if ∃(i = 1, …, ℓ) b_{j} = a_{j}, j < i and b_{i} > a_{i}. Now I_{A}(x) ⊂ I_{A}(y) implies that ρ_{A}(x) ≥ ρ_{A}(y).

5.8. Definition. For x ∈ Y_{A}^{ss} the normalisation 𝕀_{A}^{i}(x) of I_{A}^{i}(x) for 1 ≤ i ≤ m(x, A) is defined as follows. Let m := m(x, A). Then we put 𝕀_{A}^{m}(x) := I_{A}^{m}(x). If m > 1, then we define 𝕀_{A}^{i}(x) for 1 ≤ i < m inductively as follows:

𝕀_{A}^{i}(x) := ⋃{y ∈ Δ(V_{i}(x, A), R(I_{A}^{i}(x)), z)| z ∈ 𝕀_{A}^{i+1}(x)}

We will often write 𝕀_{A}(x) instead of 𝕀_{A}^{1}(x). From the definition it is clear that 𝕀_{A}(x) is the smallest polyhedron that has the same flag V_{•}(x, A), the same ”inverse volume” ρ_{A}(x) and the same center I_{A}^{m(x,A)}(x) as I_{A}(x).

§6. Group theoretical description of I_{A}(x) for split groups.

6.1. For use in §7 below, we will now give a description of the interval of semistability I_{A}(x) in more group theoretical terms. The root system Φ of G is again assumed to be of type A_{ℓ}, …, D_{ℓ} or E_{6}.

The parabolic subgroup P ⊂ G that corresponds to the highest weight λ is generated by S = T and the additive subgroups U_{α} ⊂ G with (α, λ) ≥ 0. Let x ∈ Y_{A}^{ss}. If x_{λ} ≠ 0, one can write x uniquely as x = Πu_{α}(a_{α}) · P with u_{α}(a_{α}) ∈ U_{α}, where the roots α ∈ Φ are such that (α, λ) < 0 and one has choosen a fixed order of the roots α ∈ Φ. If x is in Y_{0,A}^{ss} and after normalisation is such that |x_{λ}| = 1 then for all α ∈ Φ one has |a_{α}| ≤ 1.

Let us assume that x as above is in Y_{0,A}^{ss}. If I_{A}(x) ≠ {0}, then we assume moreover that 0 is not on a face of I_{A}(x). Then the α with |a_{α}| = 1 span V_{1}(x, A). Now for a s ∈ S(𝕂) with χ(s) = 1 for all χ ∈ V_{1}(x, A) and such that |a_{α}α(s)| ≤ 1 for all α ∈ Φ with (α, λ) < 0 one has s · x ∈ Y_{0,A}^{ss}. So s^{−1} · 0 ∈ I_{A}(x). In fact these s ∈ S(𝕂) give us all of I_{A}(x).

Since the roots α with |a_{α}| = 1, (λ, α) < 0 span V_{1}(x, A) and λ ∈ V_{1}(x, A), our conditions on s ∈ S(𝕂) are equivalent with λ(s) = 1 and |a_{α} · α(s)| ≤ 1 for the roots α with (λ, α) < 0. Moreover, |a_{α} · α(s)| ≤ 1 is equivalent to v(a_{α}) + v(α(s)) = v(a_{α}) + (α, s^{−1} · 0) ≥ 0. This in turn means that u(a_{α}) stabilises the point s^{−1} · 0 ∈ I_{A}(x) ⊂ A. This proves the following proposition:

6.2. Proposition. Let x ∈ Y_{0,A}^{ss} and λ ∈ V_{1}(x, A). We assume that x is normalised and has |x_{λ}| = 1. We fix an order on the roots α ∈ Φ with (λ, α) < 0. Let x = h · P with h = ∏_{(α,λ)<0} u_{α}(a_{α}) with |a_{α}| ≤ 1. Then I_{A}(x) consists of the points y ∈ A satisfying the following two equivalent conditions:

i) (λ, y) = 0 and v(a_{α}) + (α, y) ≥ 0 for all roots α such that (λ, α) < 0.

ii) (λ, y) = 0 and h(y) = y.

6.3. Definition. We relate the values v(a_{α}) in the expression of x as x = ∏ u_{α}(a_{α}) · P to the constants c_{α,V} defined above. Let α be a root such that α ⊥̸ V_{1}(x, A) (We do allow α ∈ V_{1}(x, A)). Then we define c_{α}^{1}(x, A) := 0 and for

1 < i ≤ m(x, A) we define c_{α}^{i}(x, A) := ∑_{j=1}^{i−1} c_{α,V_{j}(x,A)} · R(I_{A}^{j}(x)). We furthermore

put c_{α}(x, A) := c_{α}^{m}(x, A).

6.4. Lemma. Let x ∈ Y_{0,A}^{ss} and λ ∈ V_{1}(x, A). We assume that x is normalised and has |x_{λ}| = 1. We fix an order on the roots α ∈ Φ with (λ, α) < 0. Then

x = h · P with h = ∏_{(α,λ)<0} u_{α}(a_{α}). Let 1 ≤ i ≤ m(x, A). Then 0 ∈ I_{A}^{i}(x) if and

only if for all the roots α with (λ, α)) < 0 one has v(a_{α}) ≥ c_{α}^{i}(x, A).

Proof. Let 0 ∈ I_{A}^{i}(x). We put V_{j} := V_{j}(x, A) and R_{j} := R(I_{A}^{j}(x)) for j = 1, …, i− 1. For any root α ⊥̸ V_{1}(x, A), we construct a point y_{α}^{1} in I_{A}(x) such that (y_{α}^{1}, α) = c_{α}^{i}(x, A). From this it then follows that v(a_{α}) ≥ c^{i}(x, A). We put y_{α}^{i} = 0. If i > 1 then we define the points y_{α}^{j} for j = i−1, …, 1 inductively as follows. Let us assume that we have constructed the point y_{α}^{j+1} ∈ I_{A}^{j+1}(x). Since Δ(V_{j}, R_{j}, y_{α}^{j+1}) ⊂ I_{A}^{j}(x), there exists a point y_{α}^{j} ∈ I_{A}^{j}(x) such that (α, y_{α}^{j}) = (α, y_{α}^{j+1}) + c_{α,V_{j}} · R_{j}. Then (α, y_{α}^{1}) = c_{α}^{i}(x, A). Therefore v(a_{α}) ≥ c_{α}^{i}(x, A).

Now let us assume that v(a_{α}) ≥ c_{α}^{i}(x, A) for all roots α ⊥̸ V_{1}(x, A). If i = 1 then it is clear that 0 ∈ I_{A}(x). So let us assume that i > 1. Let H_{i−1} := Δ(V_{i−1}, R_{i−1}, 0). If i > 2 then we define the H_{j}, j = 1, …, i−2 inductively. Suppose H_{j+1} is defined. Then we put H_{j} := ∪{u ∈ Δ(V_{j}, R_{j}, y)| y ∈ H_{j+1}}. From the construction of the H_{j} it follows that max{|(α, y)|| y ∈ H_{1}} = c_{α}^{i}(x, A) for all roots α ⊥̸ V_{1}(x, A). Since v(a_{α}) ≥ c_{α}^{i}(x, A), it follows that u_{α}(a_{α}) stabilises H_{1}. Therefore H_{1} ⊂ I_{A}(x). From this one concludes that 0 ∈ I_{A}^{i}(x). This proves the lemma.

6.5. Lemma. Let x ∈ Y_{A}^{ss} and 1 ≤ i ≤ m(x, A). Let α ∈ Φ be a root not contained in V_{i}(x, A)^{⊥}. If i > 1, then we also assume that α ∉ V_{i−1}(x, A). Let d_{1}(x, A) := 0 and if i > 1 let d_{i}(x, A) := ∑_{j=1}^{i−1} R(I_{A}^{j}(x)) for 1 < i ≤ m(x, A). Then one has:

i) c_{α}^{i}(x, A) ≥ d_{i}(x, A).

ii) If α ∉ V_{i}(x, A)^{c}, then c_{α}^{i}(x, A) = d_{i}(x, A), except when Φ = C_{ℓ}, ℓ > 2 and

α is long. iii) c_{α}^{i}(x, A) = 2 · d_{i}(x, A), if Φ = C_{ℓ}, ℓ > 2, α ∉ V_{i}(x, A)^{c} and α is long.

Proof. This follows directly from the definitions, proposition 4.6 and lemma 4.8.

6.6. Definition. We say that a point x ∈ Y_{0,A}^{ss} is almost in standard position if λ ∈ V_{1}(x, A) and |x_{λ}| = 1 after normalisation. We say that x is in standard position if x is almost in standard position, 0 ∈ I_{A}^{m(x,A)}(x) and moreover 0 is not contained in a face of I_{A}^{m(x,A)}(x).

6.7. Lemma. Let x ∈ Y_{0,A}^{ss} be in standard position. Let m := m(x, A). Let x be x = Πu_{α}(a_{α}) · P for some fixed order of the roots α ∈ Φ with (α, λ) < 0. Then the following statements hold:

i) For all α ∈ Φ one has v(a_{α}) ≥ c_{α}(x, A)

ii) The α ∈ Φ with v(a_{α}) = c_{α}^{i}(x, A) span V_{i}(x, A).

iii) For all α ∈ Φ, α ∉ V_{m}(x, A) one has v(a_{α}) > c_{α}(x, A) = c_{α}^{m}(x, A). This can only occur if V_{m}(x, A) ≠ X (S) ⊗ ℝ, that is if x ∈ Y_{A}^{ss} − Y_{A}^{s}.

Proof. This follows directly from lemma 6.5 above.

6.8. Definition. Let x ∈ Y_{0,A}^{ss} be normalised and be such that |x_{λ}| = 1. Let m := m(x, A). We order the roots α ∈ Φ in such a way that α_{j} ∈ V_{1}(x, A) if 1 ≤ j ≤ j_{1} and if m > 1 then for 1 < i ≤ m we have α_{j} ∈ V_{i}(x, A) − V_{i−1}(x, A) if j_{i} ≥ j > j_{i−1}. Here j_{1} < j_{2} < ⋯ < j_{m}. If x ∈ Y_{A}^{ss} − Y_{A}^{s}, then α_{j} ∉ V_{m}(x, A) for j = j_{m} + 1, …, j_{m+1}. We write x = u_{α_{jm}} (a_{j_{m}}) · · · · · u_{α_{1}}(a_{1}) · P if x ∈ Y_{A}^{s}, and we write x = u_{α_{j_{m+1}}} (a_{j_{m+1}}) · · · · · u_{α_{1}}(a_{1}) · P if x ∈ Y_{A}^{ss} − Y_{A}^{s}. For 1 ≤ i ≤ m we define h_{i} := u_{α_{j_{i}}} (a_{j_{i}}) · · · · · u_{α_{1}}(a_{1}) and y_{i} := h_{i} · P.

Both h_{i} and y_{i} depend (uniquely) on x and A. It will always be clear from the context which x and A define the h_{i} and y_{i}.

6.9. Proposition. Let x ∈ Y_{0,A}^{ss} be normalised and be such that |x_{λ}| = 1 and m(x, A) > 1. Let 1 ≤ i < m(x, A) and z ∈ I_{A}^{i}(x). Then one has:

i) I_{A}(x) ⊂ I_{A}(y_{i}).

ii) If i > 1 then R(I_{A}^{j}(y_{i})) = R(I_{A}^{j}(x)) for j = 1, …, i − 1.

iii) I_{A}^{i}(y_{i}) = {s · z|s ∈ S(𝕂) χ(s) = 1, ∀(χ ∈ V_{i}(x, A))}.

Proof. After translation by a suitable element s ∈ S, we may assume that 0 ∈ I_{A}^{m(x,A)}(x). If I_{A}^{m(x,A)}(x) is not a point, then we may moreover assume that 0 is not on a face of I_{A}^{m(x,A)}(x).

Now i) follows from the fact that if s · x ∈ Y_{0,A}^{ss} then |a_{α} · α(s)| ≤ 1 for all α ∈ Φ. Hence also s · y· ∈ Y_{0,A}^{ss}. Therefore I_{A}(x) ⊂ I_{A}(y_{i}).

Next we prove statement ii). Let F_{β} be a face of I_{A}(x) corresponding to a root β ∈ V_{j}(x, A), j < i. Let s ∈ S(𝕂) be such that s^{−1} · 0 ∈ F_{β} and such that s^{−1} · 0 is not in a face of F_{β}. One easily sees that the reductions of s · x and s · y coincide. Hence the hyperplane (β, y) = c_{β}, defined by F_{β} gives also a face of I_{A}(y_{i}). From this ii) follows.

For all s ∈ S such that χ(s) = 1 for all χ ∈ V_{i}(x, A) one has sh_{i}s^{−1} = h_{i}. Hence s · y_{i} = s · h_{i}P = sh_{i}s^{−1}P ∈ Y_{0,A}^{ss}. From this iii) follows.

§7. Description of I(x) for split groups I

7.1. Introduction. Let G, X and λ be fixed. We also assume that G is split and has a root system of type A_{ℓ}, …, D_{ℓ} or E_{6}. In this section we will give a partial description of I(x).

The proofs will be mainly group theoretical. Instead of using our line bundle L we will write x as x = h · P ∈ X(𝕂) = G(𝕂)/P (𝕂). The main ingredient of the proofs will be the well known commutator relations for elements u_{α}(a) ∈ U_{α} and u_{β}(b) ∈ U_{β} (See 7.2 below). These relations will enable us to relate I_{hA}(x), for h ∈ U_{α}(K) ∩ P (𝕀_{A}^{i}(x)) with α ⊥ V_{i−1}(x, A) (if i > 1) to I_{A}(x). 7.2. Commutator relations. Let α and β ≠ ±α be roots. The groups U_{α} and U_{β} are subgroups of G that are isomorphic to K^{∘}. Let u_{α}(a) ∈ U_{α} and u_{β}(b) ∈ U_{β}. Then the commutator of these two elements is given by [u_{α}(a), u_{β}(b)] =

∏_{n,m>0} u_{nα+mβ}(d_{n,m}a^{n}b^{m}), where nα + mβ ∈ Φ and d_{n,m} ∈ K^{∘} (See [Hu] §32 and

§33 or [S] proposition 10.1.4). Though the constant d_{n,m} does depend on the roots α and β, we will not make this explicit in the notation. It will always be clear fom the context which roots are involved.

The isomorphisms between the groups U_{α} and U_{β} and K^{∘} can be fixed in such a way that all the constants d_{n,m} are integers (See [S] proposition 11.2.7 or [Hu] §33). One can choose the isomorphisms such that the only integers occurring are 0, ±1, ±2 and ±3 (See [Hu] §33). Therefore in Char(K) = 2 and 3 some exceptional commutations occur (See [BoT] 4.2 and 4.3).

7.3. Definition. Let V ⊂ X (S) ⊗ ℝ = ℝ^{ℓ} be a hyperplane spanned by roots. Then we write α ⊥⊥ V and say that α is extremely orthogonal to V if and only if the additive group U_{α} ⊂ G commutes with all the groups U_{β}, where the roots β are contained in V. Note that this in some cases also depends on the characteristic of the field K.

In the proposition below we describe in detail when α ⊥⊥ V occurs. We omit the proof, since it consists of applying the commutator relations. For the exceptional commutations occurring when Char(K) = 2 we refer to [BoT] sections 4.2 and 4.3(i) and the references given there.

7.4. Proposition. Let V ⊂ X (S)⊗ℝ be a linear subspace spanned by roots. Then α ⊥⊥ V if and only if α ⊥ V and moreover one of the following statements holds:

(1) All roots in Φ have the same length.

(2) Φ = B_{ℓ}, C_{ℓ} or F_{4} and Char(K) = 2.

(3) Φ = B_{ℓ}, C_{ℓ} or F_{4} and α is a long root.

(4) Φ = B_{ℓ} or F_{4}, α is a short root and V contains no short roots.

(5) Φ = C_{ℓ}, α is a short root and V does not contain a short root γ (unique upto sign) such that α ± γ is a long root.

(6) Φ = G_{2}.

7.5. Proposition. Let x ∈ Y_{A}^{ss} be such that m(x, A) > 1 and let 1 ≤ i < m(x, A). Let α ⊥⊥ V_{i}(x, A). Let z ∈ I_{A}^{i+1}(x) and h ∈ U_{α}(K) ∩ P_{z}. Then one has:

i) If i > 1 then V_{j}(h^{−1} · x, A) = V_{j}(x, A) and R(I_{hA}^{j}(x)) = R(I_{A}^{j}(x)) for

j = 1, …, i − 1. ii) I_{A}^{i}(x) ∩ h · A ⊂ I_{hA}^{i}(x). In particular z ∈ I_{hA}^{i}(x). iii) h(Δ(V_{i}(x, A), R(I_{A}^{i}(x)), z)) ⊂ I_{hA}^{i}(x). In particular R(I_{hA}^{i}(x)) ≥ R(I_{A}^{i}(x)).

Proof. Since α ⊥⊥ V_{i}(x, A), we have α ⊥ V_{i}(x, A). In particular the element h ∈ U_{α}(K) ∩ P_{z} does not act on the coordinates x_{w(λ)}, w ∈ W, with w(λ) ∈ V_{1}(x, A). From this and the fact that x ∈ Y_{z,A}^{ss}, one concludes that x ∈ Y_{z,hA}^{ss}. Since I_{hA}(x) = h(I_{A}(h^{−1}(x))), we only need to study I_{A}(h^{−1}(x)) to prove the lemma. We can now translate by an element s ∈ S(𝕂) such that s · z = 0. Then we have to replace h by shs^{−1}, which is not an element of U_{α}(K) anymore, but is contained in U_{α}(𝕂). In fact shs^{−1} ∈ U_{α}(𝕂^{∘}), since h ∈ U_{α}(K) ∩ P_{z}.

So let us assume that z = 0. Let h^{−1} = u_{α}(a) ∈ U_{α}(𝕂^{∘}). Then x ∈ Y_{0,A}^{ss} and 0 ∈ I_{A}^{i+1}(x). We may assume that x is normalised and has |x_{λ}| = 1. Then λ ∈ V_{1}(x, A). Since α ⊥ λ, we have h^{−1} = u_{α}(a) ∈ P. So u_{α}(a) · x = u_{α}(a)Π_{β}u_{β}(b_{β}) · P = Π_{β}(u_{α}(a)u_{β}(b_{β})u_{α}(a)^{−1}) · P. To prove the proposition we look at the expressions f_{β} := u_{α}(a)u_{β}(b_{β})u_{α}(a)^{−1}. It follows from the commutator relations for U_{α} and U_{β} (see 7.2) that f_{β} = Π_{n,m>0}u_{nα+mβ}(d_{n,m}a^{n}b_{β}^{m}) · u_{β}(b_{β}), where nα + mβ ∈ Φ and d_{n,m} ∈ K^{∘}.

Since 0 ∈ I_{A}^{i+1}(x), one has v(b_{β}) ≥ c_{β}^{i+1}(x, A) if β ∉ V_{i}(x, A). If β ∈ V_{j}(x, A), j ≤ i then v(b_{β}) ≥ c_{β}^{j}(x, A). We will show that similar statements hold for the u_{nα+mβ}(d_{n,m}a^{n}b_{β}^{m}) occurring in the f_{β}. From this the proposition will follow.

If β ∈ V_{i}(x, A) then f_{β} = u_{β}(b_{β}) and there is nothing to prove. So let us assume β ∉ V_{i}(x, A). We only have to consider β such that U_{α} and U_{β} do not commute.

First we consider the case where all the roots in Φ have the same length. If the root β is in V_{i}(x, A)^{c} then the group U_{α} commutes with U_{β}. Therefore it is enough to consider only roots β that are not in V_{i}(x, A)^{c} in this case.

Let γ be one of the roots nα +mβ with n, m > 0. Only n = m = 1 occurs. Since (β, λ) < 0 and (α, λ) = 0, one has (γ, λ) < 0. Since v(a) ≥ 0 and v(b_{β}) ≥ c_{β}^{i+1}(x, A), it follows that v(d_{1,1}ab_{β}) ≥ v(b_{β}) ≥ c_{β}^{i+1}(x, A). Since β ∉ V_{i}(x, A)^{c}, one has by lemma 6.5 that c_{β}^{i+1}(x, A) = d_{i+1}(x, A). Since α ⊥̸ γ, it is clear that γ ∉ V_{i}(x, A)^{c}. Therefore c_{γ}^{i+1}(x, A) = d_{i+1}(x, A) by lemma 6.5. Hence v(d_{1,1}ab_{β}) ≥ c_{γ}^{i+1}(x, A).

Next we consider the case where Φ = C_{ℓ}, ℓ > 2. We will assume that Char(K) ≠ 2. If Char(K) = 2, then the only difference is that some more roots commute. So the proof in that case follows easily from the case where Char(K) ≠ 2.

Let us first consider the case where the root α is short. If the root β is also short and does not commute with α, then either γ = α+β is a short root or γ = α+β is a long root. Let us first assume that γ = α + β is a short root. From γ ⊥̸ α it follows that γ ∉ V_{i}(x, A)^{c}. Therefore c_{γ}^{i+1}(x, A) = c_{β}^{i+1}(x, A) = d_{i+1}(x, A). Consequently, we have v(d_{1,1}ab_{β}) ≥ c_{γ}^{i+1}(x, A).

Now we will show that γ = α + β cannot be a long root, if α and β are short roots. Let us assume that it does occur. Let Φ consist of the roots ±2e_{i}, ±e_{i} ± e_{j}, i ≠ j, 1 ≤ i, j ≤ ℓ. Here the e_{i} form an orthonormal basis of ℝ^{ℓ}. By assumption our root β is short. So we must have β + α = γ and β ⊥ α. We may choose α = e_{1} + e_{2} and β = e_{1} − e_{2}. Since α ⊥ V_{i}(x, A), none of the roots ±2e_{1}, ±2e_{2}, ±e_{1} ± e_{j}, ±e_{2} ± e_{j}, j ≠ 1, 2 is contained in V_{i}(x, A). From this it follows that either β ⊥ V_{i}(x, A) or β ∈ V_{i}(x, A). This contradicts our assumptions on β. Therefore γ cannot be a long root.

Next we consider the case where β is a long root that does not commute with the short root α. Then γ_{1} := β + α is a short root and γ_{2} := β + 2α is a long root. For both these roots one has that c_{γ_{n}}^{i+1}(x, A) ≤ c_{β}^{i+1}(x, A), n = 1, 2. Therefore v(d_{n,1}a^{n}b_{β}) ≥ c_{γ_{n}}^{i+1}(x, A) for n = 1, 2.

Let us now assume that the root α is long. If the root β does not commute with α, then the root β is short. Then γ_{1} := α + β is a short root and γ_{2} := α + 2β is a long root. For γ_{1} one has again that c_{γ_{1}}^{i+1}(x, A) ≤ c_{β}^{i+1}(x, A). For γ_{2} one has c_{γ_{2}}^{i+1}(x, A) ≤ 2 · c_{β}^{i+1}(x, A). Hence also in this case we have v(d_{1,n}ab_{β}^{n}) ≥ c_{γ_{n}}^{i+1}(x, A) for n = 1, 2.

We will omit the proof for the case where Φ = B_{ℓ}. In this case one again proves, by looking at each possible root γ occurring in the expressions f_{β}, that v(d_{n,m}a^{n}b_{β}^{m}) ≥ c_{γ}^{i+1}(x, A) holds.

For all the roots γ = n·α+m·β occurring, we have now shown that v(d_{n,m}a^{n}b_{β}^{m}) ≥ c_{γ}^{i+1}(x, A) holds. For γ ∈ V_{i}(x, A) we have in fact shown that v(d_{n,m}a^{n}b_{β}^{m}) > c_{γ}^{i+1}(x, A). Let us rearrange the roots and write h^{−1}(x) = Π_{β}u_{β}(b′_{β})·P. For all roots β with v(b_{β}) = c_{β}^{i}(x, A) one also has that v(b′_{β}) = c_{β}^{i}(x, A). Moreover, for the roots β ∈ V_{i}(x, A) such that v(b_{β}) > c_{β}^{i}(x, A) one has that v(b′_{β}) > c_{β}^{i}(x, A). Furthermore for the roots α ∉ V_{i}(x, A) one has that v(b′_{β}) ≥ c_{β}^{i+1}(x, A). From this one concludes that if i > 1 then V_{j}(h^{−1}(x), A) = V_{j}(x, A) and R(I_{A}^{j}(h^{−1}(x))) = R(I_{A}^{j}(x)) for j = 1, …, i − 1. This proves (i).

Let y ∈ I_{A}^{i}(x) ∩ h · A. Let s ∈ S(𝕂) be such that s · y = 0. Then s · x = Π_{β}u_{β}(b_{β} · β(s)) · P. Moreover s · h^{−1} · x = Π_{β}(b′_{β} · β(s)) · P. Since y ∈ I_{A}^{i}(x), we have that v(b_{β} · β(s)) ≥ c_{β}^{i}(x, A). The same arguments as above used to show that v(b′_{β}) ≥ c_{β}^{i+1}(x, A) give us that v(b′_{β} · β(s)) ≥ c_{β}^{i}(x, A). From this one concludes that y ∈ I_{hA}^{i}(x). This proves (ii).

From (ii) it follows that z ∈ I_{A}^{i}(h^{−1}(x)). For all β ∉ V_{i}(x, A) one has that v(b′_{β}) ≥ c_{β}^{i+1}(x, A). From this one concludes that Δ(V_{i}(x, A), R(I_{A}^{i}(x), 0) ⊂ I_{A}^{i}(h^{−1}(x)). This proves (iii).

7.6. Proposition. Let x ∈ Y_{A}^{ss} and m(x, A) > 1. Let 1 ≤ i < m(x, A). Let α ⊥⊥ V_{i}(x, A) and h ∈ U_{α}(K) ∩ P (𝕀_{A}^{i+1}(x)). Then h(𝕀_{A}(x)) ⊂ I_{hA}(x).

Proof. This follows from the previous proposition.

7.7. Definition. Let x ∈ Y_{A}^{ss} and 1 ≤ i < m(x, A). Then we write α ⊥⊥_{x} V_{i}(x, A) if and only if α ⊥ V_{i}(x, A) and for all h ∈ U_{α}(𝕂) ∩ P (𝕀_{A}^{i+1}(x))_{𝕂} one has that 𝕀_{A}(x) ⊂ I_{A}(h^{−1} · x). Here P (𝕀_{A}^{i+1}(x))_{𝕂} denotes the stabiliser of 𝕀_{A}^{i+1}(x) in G(𝕂).

Note that α ⊥⊥ V_{i}(x, A) implies that α ⊥⊥_{x} V_{i}(x, A). Below we determine for which short roots α ∈ Φ one has α ⊥⊥_{x} V_{i}(x, A).

7.8. Proposition. Let Φ = B_{ℓ}, ℓ ≥ 2, Char(K) ≠ 2 and x ∈ Y_{0,A}^{ss} be such that m(x, A) > 1. Let 1 ≤ i < m(x, A). We assume that x is almost in standard position and that 0 ∈ I_{A}^{i+1}(x). Let x = Πu_{β}(b_{β}) · P. Let α ⊥ V_{i}(x, A) be a short root.

Then α ⊥⊥_{x} V_{i}(x, A) if and only if for all short roots β ∈ V_{i}(x, A) one has that

v(2 · b_{β}) ≥ ∑_{j≤i} R(I_{A}^{j}(x)).

Proof. From the fact that α ⊥ V_{1}(x, A) and x ∈ Y_{0,A}^{ss} , it follows that h^{−1} · x ∈ Y_{0,A}^{ss}. Let h^{−1} = u_{α}(a) ∈ U_{α}(𝕂^{∘}). To prove the proposition we again look at the expressions f_{β} in u_{α}(a)·x = u_{α}(a) ∏_{β} u_{β}(b_{β})·P. For all the long roots β ∈ V_{i}(x, A) we have f_{β} = u_{β}(b_{β}). For all the f_{β}, β ∉ V_{i}(x, A) one has again that in the u_{nα+mβ}(d_{n,m}a^{n}b_{β}^{m}) occurring in f_{β} the inequalities v(d_{n,m}a^{n}b_{β}^{m}) ≥ c_{nα+mβ}^{i+1}(x, A) hold.

So we only have to look at the f_{β} with β short, (β, λ) < 0 and β ∈ V_{i}(x, A). One has f_{β} = u_{α+β}(d_{1,1}ab_{β})u_{β}(b_{β}). If i > 1 then one concludes from the fact that α ⊥⊥_{x} V_{i−1}(x, A) that v(d_{1,1}ab_{β}) ≥ c_{β+α}^{i}(x, A). Moreover this is true for all the short roots β contained in V_{i}(x, A). We only have to consider a short root β such that the value v(b_{β}) is minimal. Furthermore we may choose h in such a way that the value v(a) is minimal, that is v(a) = 0.

If v(d_{1,1}ab_{β}) ≥ c_{β+α}^{i+1}(x, A) then statements i), ii) and iii) of proposition 7.5 hold. From this one concludes as in proposition 7.6 that one has α ⊥⊥_{x} V_{i}(x, A). If v(d_{1,1}ab_{β}) < c_{β+α}^{i+1}(x, A) then Δ(V_{i}(x, A), R(I_{A}^{i}(x)), z) is not contained in I_{A}^{i}(h^{−1} · x). Hence one cannot have α ⊥⊥_{x} V_{i}(x, A). (Note that v(d_{1,1}) = v(2) ).

7.9. Remark. Next we turn to the root system C_{ℓ}. If α ∈ Φ is a short root there exist a short root γ, unique up to sign, such that α + γ and α − γ are long roots. Moreover α ⊥ γ.

Let V be a linear subspace of X (S) ⊗ ℝ ≃ ℝ^{ℓ} spanned by roots. Let α ⊥ V. If γ ∈ V then V = V′⊕ < γ > with γ ⊥ V′. Now suppose V = V_{i}(x, A) for some x ∈ Y_{A}^{ss}. If γ ∈ V_{i}(x, A) then it follows from proposition 5.4 that γ ∈ V_{1}(x, A).

7.10. Proposition. Let Φ = C_{ℓ}, ℓ ≥ 3 and Char(K) ≠ 2. Let x ∈ Y_{A}^{ss} be such that m(x, A) > 1. Let 1 ≤ i < m(x, A). Let α ⊥ V_{i}(x, A) be a short root and γ ∈ Φ be a short root (unique upto sign) such that α + γ is a long root. We assume that γ ∈ V_{i}(x, A). Let z ∈ I_{A}^{i+1}(x) and h ∈ U_{α}(K) ∩ P_{z}. Then the following holds:

i) If K is not a finite extension of ℚ_{2} then I_{hA}(x) = I_{A}(x)∩hA. In particular

one never has α ⊥⊥_{x} V_{1}(x, A).
ii) If K ⊃ ℚ_{2} then α ⊥⊥_{x} V_{i}(x, A) if and only if v(2) ≥ \frac{1}{2} ∑_{j=1}^{i} R(I_{A}^{j}(x)).

Proof. From α ⊥ V_{1}(x, A) and x ∈ Y_{z,A}^{ss}, it follows that x ∈ Y_{z,hA}^{ss}. After translation by an element s ∈ S(𝕂), we may again assume that z = 0. We may moreover assume that after normalisation one has |x_{λ}| = 1. Now one considers u_{α}(a) · x = u_{α}(a) · Π_{β}u_{β}(b_{β}) · P, h^{−1} = u_{α}(a) and β such that (β, λ) < 0. Again one looks at the expressions f_{β}. The only such expression different from the ones occurring in the proof of proposition 7.5 is f_{γ}. So we will only consider this expression.

Now let us assume that γ ∈ V_{i}(x, A). After replacing γ by −γ, if necessary, we may assume that (γ, λ) < 0. By remark 7.9 one has that γ ∈ V_{1}(x, A) and V_{1}(x, A) = V′⊕ < γ > with γ ⊥ V′. Then one has f_{γ} = u_{α+γ}(d_{1,1}ab_{γ})u_{γ}(b_{γ}) with v(b_{γ}) = 0. From [BoT] 4.2 it follows that d_{1,1} is a unit if K is not a finite extension of ℚ_{2} and that v(d_{1,1}) = v(2) > 0 if K ⊃ ℚ_{2}.

Hence if K does not contain ℚ_{2} then v(d_{1,1}ab_{γ}) = 0 and I_{hA}(x) = I_{A}(x) ∩ hA. This proves i). If K ⊃ ℚ_{2} then v(d_{1,1}ab_{γ}) = v(2). Moreover c_{β+γ}^{i+1}(x, A) =

2 ∑_{j=1}^{i} R(I_{A}^{j}(x)) and c_{β+γ}^{i}(x, A) = 2 ∑_{j=1}^{i−1} R(I_{A}^{j}(x)). From this ii) follows in a way similar to the proof of proposition 7.8. 7.11. Proposition. Let x ∈ Y_{A}^{ss} be such that m(x, A) > 1. Let 1 ≤ i < m(x, A). Let α ⊥ V_{i}(x, A) be a short root. We assume that not α ⊥⊥_{x} V_{i}(x, A). If i > 1, then we assume that α ⊥⊥_{x} V_{i−1}(x, A). Let z ∈ I_{A}^{i+1}(x).

Then there exists an element s ∈ S(𝕂) such that for all h ∈ U_{α}∩P_{z}, the translate H_{h} := s·(A∩hA) of A∩hA with A∩hA ⊂ H_{h} is such that h(Δ(V_{i}(x, A), R(I_{A}^{i}(x)), y)∩ H_{h}) = h(Δ(V_{i}(x, A), R(I_{A}^{i}(x)), y)) ∩ I_{hA}^{i}(x). Here s only depends on x and α. Proof. The root system Φ is of type B_{ℓ} or of type C_{ℓ}. The case with Φ = C_{ℓ} and the characteristic of the residue field of K unequal to two has been treated in proposition 7.10 (i). Note that in that case one always can take s = 1. So we only have to prove the proposition for the case Φ = C_{ℓ}, K ⊃ ℚ_{2} and the case Φ = B_{ℓ}. The proofs in both cases are rather similar. Therefore we will only treat the case where Φ = B_{ℓ} (and Char(K) ≠ 2).

We may assume that z = 0 and that x = ∏_{β} u_{β}(b_{β}) · P is in standard position.

As in the proof of proposition 7.8, we may assume that h^{−1} = u_{α}(a) with v(a) = 0. Let h^{−1}(x) = u_{α}(a) ∏_{β} u_{β}(b_{β})·P = ∏_{β} u_{β}(b′_{β})·P. From the proof of proposition 7.8

it follows that v(b′_{β}) ≥ c_{β}^{i+1}(x, A) for all roots β that are not of the form β = α + γ. Here γ is a short root contained in V_{i}(x, A). Let γ_{0} be a short root such that v(b_{γ_{0}}) is minimal. Then c_{α+γ_{0}}^{i}(x, A) ≤ v(b′_{γ_{0}}) = v(2ab_{γ_{0}}) < c_{α+γ_{0}}^{i+1}(x, A). For the other short roots γ in V_{i}(x, A) one has v(b′_{α+γ}) ≥ v(b′_{α+γ_{0}}).

Since α + γ_{0} ≡ α mod V_{i}(x, A), it follows from the values of the v(b′_{β}) that Δ(V_{i}(x, A), R(I_{A}^{i}(x)), z) ∩ s · H_{h} is contained in I_{A}^{i}(h^{−1} · x) for a suitably choosen s ∈ S(𝕂). From the fact that s ∈ S(𝕂) only depends on the value v(2 · b_{γ_{0}}) and not on v(a), one concludes that s only depends on x and α. 7.12. Remark. If Φ = B_{ℓ} then the statement of the proposition above is more or less independent of the short root α ⊥ V_{i}(x, A) chosen. Indeed let γ be another short root with γ ⊥ V_{i}(x, A). Let g ∈ U_{γ} ∩ P_{z}. Again one may assume that z = 0 and that z is on the wall bounding A ∩ gA. Let w ∈ W be such that w(α) = γ. Then if not α ⊥⊥_{x} V_{i}(x, A) then also not γ ⊥⊥_{x} V_{i}(x, A). The corollary above then holds for γ with s replaced by wsw^{−1}. 7.13 Definition. Let x ∈ Y_{A}^{ss}. Let 1 ≤ i ≤ m(x, A). Let α ⊥ V_{1}(x, A). Then we define U_{(α,x)} ⊂ U_{α}(K) as being the subgroup consisting of the elements h ∈ U_{α} ∩ P (𝕀_{A}^{m(x,A)}(x)) such that h(𝕀_{A}(x)) ⊂ I_{hA}(x).

Let F_{i}(x, A) be the group generated by 1 and the groups U_{(α,x)} with α ∈ Φ such that α ⊥ V_{i}(x, A), not α ⊥⊥_{x} V_{i}(x, A) and if i > 1 then α ⊥⊥_{x} V_{i−1}(x, A). It follows from the definition that the roots α occurring are short roots. Moreover from the results obtained above one deduces that for at most one i one has F_{i}(x, A) ≠ {1}.

Let G_{i}(x, A) be the following group:

G_{1}(x, A) :=P (𝕀_{A}^{1}(x)).
G_{i}(x, A) :=P (𝕀_{A}^{i}(x))∩ < U_{α}|α ⊥⊥_{x} V_{i−1}(x, A) > if i > 1 .

Here 𝕀_{A}^{i}(x) is the normalization of I_{A}^{i}(x). We denote by H_{i}(x, A) the group generated by F_{i}(x, A) and G_{i}(x, A). We define the group H(x, A) as the group generated by the groups H_{i}(x, A), i = 1, …, m(x, A). 7.14. Definition. We call a triple (Φ, K, Y_{A}^{ss}) disturbed if there exists a point x ∈ Y_{A}^{ss} with m(x, A) > 1, such that there exists a root α ∈ Φ and 1 ≤ i < m(x, A) with α ⊥⊥_{x} V_{i}(x, A) but not α ⊥⊥ V_{i}(x, A).

Since this definition does not depend on the choice of the department A, we will also call (Φ, K, Y^{ss}) disturbed if (Φ, K, Y_{A}^{ss}) is disturbed for some apartment A.

Let us restrict ourselves to the root systems Φ = A_{ℓ}, …, D_{ℓ} and E_{6} as usual. Then disturbed triples only exist if (Φ, K) is such that either Φ = B_{ℓ}, ℓ ≥ 3 and Char(K) ≠ 2 or Φ = C_{ℓ}, ℓ ≥ 2 and K ⊃ ℚ_{2}. The following proposition is a direct consequence of the definitions and the results obtained sofar.

7.15. Proposition. Let x ∈ Y_{A}^{ss} and 1 ≤ i ≤ m(x, A). Then the following holds: i) For at most one i one has H_{i}(x, A) ≠ G_{i}(x, A). ii) If (Φ, K, Y_{A}^{ss}) is not disturbed, then H_{i}(x, A) = G_{i}(x, A) for all i.

7.16 Lemma. Let x ∈ Y_{A}^{ss} with m(x, A) > 1. Let 1 ≤ n < m ≤ m(x, A). Then the group H_{m}(x, A) normalises the group of H_{n}(x, A).

Proof. The groups H_{i}(x, A) are generated by the subgroups U_{γ} ∩ H_{i}(x, A). Let u_{β}(b) ∈ H_{n}(x, A) and u_{α}(a) ∈ H_{m}(x, A). To prove the lemma it is sufficient to prove that f_{β} := u_{α}(a)u_{β}(b)u_{α}(a)^{−1} ∈ H_{n}(x, A).

Let us first assume that n = 1 and that H_{1}(x, A) = G_{1}(x, A). There exist points y := ∏_{γ} u_{γ}(a_{γ}) · P such that I_{A}(y) = 𝕀_{A}(x) and H_{j}(y, A) = H_{j}(x, A), j = 1, …, m(x, A). After replacing λ by w(λ) (or β by w(β) ), where w is a suitable element in the Weyl group of Φ ∩ V_{1}(y, A), we may assume that (β, λ) < 0. Let h := u_{α}(a).

In the proofs of the propositions above (especially 7.5, 7.8 and 7.10), we have in

fact shown that f_{β} = hu_{β}(b_{β})h^{−1} = ∏_{δ} u_{δ}(d_{δ}) · u_{β}(b_{β}), δ = i · α + j · β, i, j > 0,

where the u_{δ}(d_{δ}) are such that they stabilise 𝕀_{A}(y) = 𝕀_{A}(x). Therefore f_{β} ∈ P (𝕀_{A}(y)) = H_{1}(y, A) = H_{1}(x, A). Since this holds for all groups U_{β} ∩ H_{1}(x, A), it follows that the group H_{m}(x, A) normalises H_{1}(x, A).

Let us now consider the case where n = 1 and H_{1}(x, A) ≠ G_{1}(x, A). One can still construct points y ∈ Y_{A}^{ss} such that I_{A}(y) = 𝕀_{A}(x) and H_{j}(y, A) = H_{j}(x, A) for j = 1, …, m(x, A). We may again assume that (β, λ) < 0 and that y = ∏ u_{γ}(a_{γ}) · P. If u_{β}(b) ∈ G_{1}(x, A), then the argument used above for the case where H_{1}(x, A) = G_{1}(x, A) applies and we are done.

So let us assume that u_{β}(b) ∉ G_{1}(x, A). It is sufficient to prove the lemma for the generators of F_{1}(x, A). Therefore we may assume that β is a short root, such that β ⊥ V_{1}(x, A) = V_{1}(y, A) but not β ⊥⊥_{y} V_{1}(y, A). From proposition 7.15 and definition 7.14 it follows that we only have to consider the case where Φ = B_{ℓ}, ℓ ≥ 3 and Char(K) ≠ 2 and the case where Φ = C_{ℓ}, ℓ ≥ 2 and K ⊃ ℚ_{2}.

Let us first consider the case where Φ = B_{ℓ}, ℓ ≥ 3 and Char(K) ≠ 2. Now the root α must be long, since there are no short roots γ such that γ ⊥⊥_{y} V_{1}(y, A). If (α, β) ≥ 0, then f_{β} = u_{β}(b) and we are done. So let us assume that (α, β) = −1. Then f_{β} = u_{α+β}(d_{1,1}ab) · u_{α+2β}(d_{1,2}ab^{2}) · u_{β}(b). Each of the terms occurring in the expression f_{β} is in H_{1}(y, A) (in fact in F_{1}(x, A)). So the lemma holds in this case.

Let us now consider the case where Φ = C_{ℓ}, ℓ ≥ 2 and K ⊃ ℚ_{2}. Again we only have to consider short roots β ⊥ V_{1}(y, A) such that not β ⊥⊥_{y} V_{1}(y, A). One now easily sees that for all roots α ⊥⊥_{y} V_{1}(y, A) the groups U_{α} and U_{β} commute. Hence f_{β} = u_{β}(b) ∈ H_{1}(y, A) = H_{1}(x, A). So the lemma also holds in this case.

Now we look at the case where n > 1. One can now construct a point y ∈ Y_{A}^{ss} such that I_{A}(y) = 𝕀_{A}^{n}(x), H_{i}(y, A) = H_{i+n−1}(x, A) for i = 2, …, m(x, A) − n + 1 and H_{n}(x, A) = H_{1}(y, A)∩ < U_{α} | α ⊥⊥_{x} V_{n−1}(x, A) >. Now one can apply the proof used for the case where n = 1 by replacing x by y and only considering the roots β such that β ⊥⊥ V_{n−1}(x, A). This concludes the proof of the lemma.

7.17. Theorem. Let x ∈ Y_{A}^{ss}. Let h ∈ H(x, A) be such that r_{h·A,A}(x) = 1. Then h(𝕀_{A}(x)) ⊂ I_{hA}(x).

Proof. Let us first assume that h ∈ H_{i}(x, A) for some i = 1, …, m(x, A). If h = u_{α}(a), then it follows from the definition of H_{i}(x, A). So let us assume that h = ∏_{α} u_{α}(a_{α}). If i > 1, then it follows from the proofs that V_{j}(u_{α}(a_{α}) · x, A) = V_{j}(x, A) and R(I_{A}^{j}(u_{α}(a_{α}) · x)) = R(I_{A}^{j}(x)) for j = 1, …, i − 1. Moreover one has that 𝕀_{A}(x) ⊂ I_{A}(u_{α}(a_{α}) · x). From this one concludes that 𝕀_{A}(x) ⊂ I_{A}(h^{−1} · x).

Let us now assume that h is not contained in H_{i}(x, A) for any i = 1, …, m(x, A). Then we can write h as h = h_{m} ⋯ h_{1} with h_{i} ∈ H_{i}(x, A). We again look at I_{A}(h^{−1}x) instead of I_{hA}(x). For i = 1, …, m := m(x, A) one has V_{j}(h_{i}^{−1} ⋯ h_{m}^{−1} · x, A) = V_{j}(x, A) and R(I_{A}^{j}(h_{i}^{−1} ⋯ h_{m}^{−1} · x)) = R(I_{A}^{j}(x)) for j = 1, …, i. Moreover, 𝕀_{A}^{i}(x) ⊂ I_{A}^{i}(h_{i}^{−1} ⋯ h_{m}^{−1}·x). Hence 𝕀_{A}(x) ⊂ I_{A}(h_{i}^{−1} ⋯ h_{m}^{−1}·x) for all i = 1, …, m := m(x, A). This proves the lemma.

7.18. Remark. The condition r_{h·A,A}(x) = 1 in the theorem above is always satisfied if A ∩ I(x) ≠ ∅ or if h = h_{m} ⋯ h_{j+1} · h_{j}, h_{i} ∈ H_{i}(x, A) with j > 1.

7.19 Definition. We define a function ρ : Y^{ss} ⟶ ℝ^{ℓ} that measures the maximal volume of I_{A}(x) for x ∈ Y^{ss}. Here ℓ is the rank of G. On ℝ^{ℓ} we put the lexicographical ordering. The apartment A with ρ_{A}(x) minimal is the apartment where I_{A}(x) is maximal. We define:

ρ(x) := inf {ρ_{A}(x)|A ∩ I(x) ≠ ∅} .

We say that A determines I(x) if A ∩ I(x) ≠ ∅ and ρ_{A}(x) = ρ(x).

§8. Description of I(x) for split groups II

8.1. Introduction. We continue our description of I(x) for split groups G with root system Φ of type A_{ℓ}, …, D_{ℓ} or E_{6}.

8.2. Lemma. Let x = ∏_{β} u_{β}(b_{β}) · P ∈ Y_{A}^{ss}, (β, λ) < 0 and 1 ≤ i < m(x, A). Let α ∉ V_{i}(x, A) be a root such that α ⊥̸ V_{i}(x, A) (and α ⊥⊥_{x} V_{i−1}(x, A) if i > 1). We assume that 0 ∈ I_{A}^{i+1}(x). Then one of the following statements holds:

i) (λ, α) < 0 (and then i = 1).

ii) There exists a root β ∈ V_{i}(x, A), (β, λ) < 0 such that (β, α) < 0 and

v(b_{β}) = c_{β}^{i}(x, A) (and β ∉ V_{i−1}(x, A) if i > 1).

Proof. Let Ψ := {β ∈ Φ|β ∈ V_{i}(x, A), (β, λ) < 0 and v(b_{β}) = c_{β}^{i}(x, A)}. If i = 1 then the β ∈ Ψ span V_{1}(x, A), otherwise the β ∈ Ψ span V_{i}(x, A) mod V_{i−1}(x, A).

First we consider the case where (λ, α) ≠ 0. We will show that if statement ii) is not true, then statement i) holds. Since (λ, α) ≠ 0 and λ ∈ V_{1}(x, A) we must have i = 1. Let us assume that for all β ∈ Ψ one has (β, α) ≥ 0. Since x ∈ Y_{0,A}^{ss} one has that 0 is in the cone λ + ∑ ℝ_{≥0} · β, β ∈ Ψ (see [PV] proposition 4.10). Hence there exist n_{β} ≤ 0 such that λ = ∑ n_{β} · β. Therefore (λ, α) < 0 and i) holds.

Next we consider the case where (λ, α) = 0. Again let us assume that (β, α) ≥ 0 for all β ∈ Ψ. By duality α defines a one parameter subgroup of S. We choose an element s of this 1 − ps such that v(β(s)) ≥ 0 for all β ∈ Ψ. Since v(b_{γ}) ≥ c_{γ}^{i+1}(x, A) > c_{γ}^{i}(x, A) for all roots γ ∉ V_{i}(x, A), (γ, λ) < 0, we can choose s in such a way that v(γ(s)b_{γ}) > c_{γ}^{i}(x, A) for all γ ∉ V_{i}(x, A).

Then s · x ∈ Y_{0,A}^{ss}. Clearly Ψ′ := {β ∈ Φ|β ∉ V_{i−1}(x, A), (β, λ) < 0 and v(b_{β} · β(s)) = c_{β}^{i}(x, A)} spans a proper (possibly empty) subset of V_{i}(x, A) (mod V_{i−1}(x, A) if i > 1). From I_{A}(s · x) = s · I_{A}(x), it follows that < Ψ′, V_{i−1}(x, A) > contains V_{i}(s · x, A) = V_{i}(x, A). This cannot be. Hence there must exist a root β ∈ Ψ such that (β, α) < 0. This proves that statement ii) holds if (λ, α) = 0.

8.3. Lemma. Let x ∈ Y_{A}^{ss} and 1 ≤ i < m(x, A). Let α be a root such that α ∉ V_{i}(x, A) and α ⊥̸ V_{i}(x, A). If i > 1 then we moreover assume that α ⊥⊥_{x} V_{i−1}(x, A). If i = 1 then we assume that A ∩ I(x) ≠ ∅. Let z ∈ I_{A}^{i}(x) and h ∈ U_{α} ∩ P_{z}. We assume that h ∉ P (I_{A}^{i}(x)). Then I_{hA}^{i}(x) = I_{A}^{i}(x) ∩ hA. If i > 1 then R(I_{hA}^{j}(x)) = R(I_{A}^{j}(x)) for j = 1, …, i − 1.

Proof. We may assume that z is on the boundary of A ∩ hA. Instead of considering I_{hA}(x), we will look at I_{A}(h^{−1}(x)). We may moreover (after possibly translating by an element s ∈ S(𝕂)) assume that z = 0. Let h^{−1} := u_{α}(a) ∈ U_{α}(𝕂^{∘}). Since 0 is on the boundary of A∩hA, one has |a| = 1. We may assume that x is normalised and has |x_{λ}| = 1. Now we again write x as x = ∏_{β} u_{β}(b_{β}) · P.

If i = 1, then it follows from A∩I(x) ≠ ∅, that I_{hA}(x) = I(x)∩hA ⊃ I_{A}(x)∩hA. Similarly for i > 1, it follows from α ⊥⊥_{x} V_{i−1}(x, A), that I_{hA}^{i−1}(x) ⊃ I_{A}^{i−1}(x) ∩ hA. Moreover, the condition that R(I_{A}^{i}(x)) = R(I_{hA}^{i}(x)) ensures us that z ∈ I_{hA}^{i}(x). Therefore we have in fact that I_{A}^{i}(x) ∩ h · A ⊂ I_{hA}^{i}(x). Hence we only have to prove that the point z = 0 is on the boundary of I_{hA}^{i}(x).

First we assume that (λ, α) ≠ 0. Then i = 1. If (λ, α) < 0, then one easily sees that our point 0 is on the boundary of I_{A}(h^{−1}·x), where h^{−1}·x = u_{α}(a) ∏_{β} u_{β}(b_{β})·P.

Next we consider the case where (λ, α) > 0. Note that in this case u_{α}(a) ∈ P. There exists for the roots β ∈ V_{1}(x, A) with v(b_{β}) = 0 positive real numbers n_{β} such that Σn_{β} · β + λ = 0. Hence some root β ∈ V_{1}(x, A) with v(b_{β}) = 0 satisfies (β, α) < 0. Again one considers the expressions f_{β} := u_{α}(a)u_{β}(b_{β})u_{α}(a)^{−1}. For our root β with (β, α) < 0, we have that γ := α + β is a root. Clearly v(d_{1,1}ab_{β}) = 0. From this one concludes that our point z = 0 is on the boundary of I_{A}(h^{−1}x).

Next we consider the case where i = 1 and (λ, α) = 0. Again there exists a root β ∈ V_{1}(x, A) with v(b_{β}) = 0, such that (β, α) < 0. From this one again concludes by considering the expressions f_{β} that z = 0 is on the boundary of I_{A}(h^{−1}x).

Finally we consider the case where i > 1. Since α ⊥ V_{i−1}(x, A), the roots β ∈ V_{i}(x, A) with (β, α) ≠ 0 are such that β ∉ V_{i−1}(x, A)^{c}. Hence c_{β}^{i}(x, A) = d_{i}(x, A), unless Φ = C_{ℓ}, ℓ > 2 and the root β is long. In that case we have c_{β}^{i}(x, A) = 2 · d_{i}(x, A). Now we again consider the expressions f_{β} := u_{α}(a)u_{β}(b_{β})u_{α}(a)^{−1} = ∏_{nα+mβ} u_{nα+mβ}(d_{n,m}a^{n}b_{β}^{m}) · u_{β}(b_{β}). We consider the expression f_{β} for a root β ∈

V_{i}(x, A), such that v(b_{β}) = c_{β}^{i}(x, A).

Let us first consider the case where Φ ≠ C_{ℓ}, ℓ > 2. Then v(b_{β}) = d_{i}(x, A). We look at the root γ := α + β that occurs in the expression f_{β}. The root γ has the same length as β. Hence c_{γ}^{i}(x, A) = d_{i}(x, A). Clearly v(d_{1,1}ab_{β}) = c_{γ}^{i}(x, A). From this we conclude that our point z = 0 is on the boundary of I_{A}^{i}(h^{−1}x).

Let us now consider the case where Φ = C_{ℓ}, ℓ > 2. We now choose our root γ occurring in the expression f_{β} as follows. If both α and β are short then we take γ := α + β, if α is long and β is short then we take γ := α + 2 · β and if α is short and β is long then we take γ := α + β. One verifies that in all these cases v(d_{n,m}a^{n}b_{β}^{m}) = m · v(b_{β}) = m · c_{β}^{i}(x, A) = c_{γ}^{i}(x, A). From this it follows that also in this case our point z = 0 is on the boundary of I_{A}^{i}(h^{−1}x). 8.4. Definition. Let x ∈ Y_{A}^{ss} be such that m(x, A) > 1. Let 1 ≤ i < m(x, A).

Let α ∈ Φ be a root such that α ∉ V_{i}(x, A) and α ⊥̸ V_{i}(x, A). Let Φ_{α}^{i}(x, A) := Φ∩ < V_{i}(x, A), α >. We define the subset Ψ_{α}^{i}(x, A) as being the set Ψ_{α}^{i}(x, A) := {β ∈ Φ| ∃(n > 0) β = n · α mod V_{i}(x, A)}. Then Ψ_{α}^{i}(x, A) ⊂ Φ_{α}^{i}(x, A). By H_{α}^{i}(x, A) we will denote the group H_{α}^{i}(x, A) :=< U_{β}|β ∈ Ψ_{α}^{i}(x, A) >. Before describing I_{hA}(x), h ∈ H_{α}^{i}(x, A) we need some lemmas. 8.5. Lemma. Let the notation be as in definition 8.4 above.

i) Then H_{α}^{i}(x, A) = ∏_{β} U_{β} for any fixed order of the roots β ∈ Ψ_{α}^{i}(x, A).

ii) If Ψ_{α}^{i}(x, A) contains a root γ ⊥ V_{i}(x, A), then γ is unique and the group U_{γ} commutes with all the groups U_{β} with β ∈ Ψ_{α}^{i}(x, A).

Proof. We first prove i). Let H_{i,α} :=< U_{β}| β ∈ Φ_{α}^{i}(x, A) >. Let μ ∈< V_{i}(x, A), α > be a weight such that μ ⊥ V_{i}(x, A) and (μ, α) > 0. Let P_{i,α} be the maximal parabolic subgroup of H_{i,α} corresponding to the weight μ. Then P_{i,α} is generated by the groups U_{β} with (β, μ) ≥ 0 and the connected component of the intersection S ∩ H_{i,α} that contains the identity.

From the definition it follows that H_{α}^{i}(x, A) is the unipotent radical of P_{i,α}. In particular it is clear that i) holds.

Let us assume that the set Ψ_{α}^{i}(x, A) contains a root γ ⊥ V_{i}(x, A). It is clear that γ must be unique. One verifies that (β, γ) > 0 for all roots β ∈ Ψ_{α}^{i}(x, A). From this it follows that the group U_{γ} commutes with the group U_{β} for any β ∈ Ψ_{α}^{i}(x, A).

8.6. Remark. Let us fix a simple basis of the root system Φ_{α}^{i}(x, A) such that the linear space generated by the roots in Φ_{α}^{i}(x, A) ∩ V_{i}(x, A) is spanned by simple roots. Let us moreover assume that the root α is such that there exists a root γ in Φ_{α}^{i}(x, A) that is orthogonal to V_{i}(x, A). Then γ is a fundamental weight for our simple basis. In particular γ is the highest long or the highest short root in the irreducible component of Φ_{α}^{i}(x, A) that contains α. Moreover this irreducible component cannot be of type A_{j}, j ≥ 1.

8.7. Proposition. Let x ∈ Y_{A}^{ss} be such that m(x, A) > 1. Let 1 ≤ i < m(x, A). Let A be such that ρ_{A}(x) = ρ(x). Let z ∈ I_{A}^{i}(x). Let h ∈ H_{α}^{i}(x, A) ∩ P_{z} be such that h ∉ P (I_{A}^{i}(x)) and such that I_{hA}^{i}(x) ≠ I_{A}^{i}(x) ∩ hA.

Then there exists a point y ∈ Y^{ss} with the following properties:

i) ρ_{A}(y) = ρ(y).

ii) V_{j}(y, A) = V_{j}(x, A) for j = 1, …, i.

iii) If i > 1 then R(I_{A}^{j}(y)) = R(I_{A}^{j}(x)) for j = 1, …, i − 1.

iv) I_{A}^{i}(y) = {s · z | s ∈ S(𝕂), χ(s) = 1, ∀χ ∈ V_{i}(x, A)}.

v) h(y) = y.

Proof. Note that it follows from lemma 8.3 above that one always has that I_{A}^{i}(x) ∩ h · A ⊂ I_{A}^{i}(x). We consider the point y_{i} associated to x and A as defined in 6.8 above.

By construction the point y_{i} has properties ii), iii) and iv). Moreover it is clear that in our case the point y_{i} also satifies property i).

Let us first consider the case where i = 1. Without loss of generality we may assume that z is on the boundary of A ∩ hA. We look at the reduction of y_{i} at z (If z = 0 then the reduction \overline{y_{i}} ∈ X ⊗ \overline{𝕂^{∘}}, where \overline{𝕂^{∘}} is the residue field. Otherwise one defines this by translating z to 0 by an element of S(𝕂).). Let Z be the variety consisting of the points x ∈ X with x_{λ} = 0 for all λ ∉ V_{1}(x, A). Let Z^{∘} be the connected component of Z that contains y_{i}. Let Z_{h}^{∘} be the variety consisting of the points in Z^{∘} that are fixed by h. Then Z_{h}^{∘} is a proper variety defined over 𝕂^{∘}.

By construction y_{i} = g(x) for some g ∈ P (I_{A}(x))_{𝕂}. From this one concludes that y_{i} also has the property that I_{hA}^{i}(y_{i}) ≠ I_{A}^{i}(y_{i}) ∩ hA. Hence the reduction of y_{i} is in the reduction of Z_{h}^{∘}. Therefore there exists a point y ∈ Z_{h}^{∘} that has the same reduction as y_{i}. One verifies that this point y has all the required properties.

If i > 1, then the construction of the point y is basically the same. The only difference is that one has to define the variety Z_{h}^{∘} in a different way over 𝕂^{∘} (without changing Z_{h}^{∘} ⊗ 𝕂). The change is such that one can lift the reduction of y_{i} to a point y with the required properties. We omit the details.

8.8. Proposition. Let x, y, A and h be as in proposition 8.7 above. Then the unipotent element h is contained in a semisimple group H defined over K that has K-rank one and stabilises y.

Proof. To proof the proposition we consider the stabiliser of the point y in the group H_{i,α} :=< U_{β}| β ∈ Φ_{α}^{i}(x, A) >. Since the coordinates y_{λ} of y are zero for all λ ∉ V_{i}(x, A), we may replace X by the subvariety Z on which x_{λ} = 0 for all λ ∉< V_{i}(x, A), α >. We moreover only consider the connected component Z^{∘} that contains y. Then Z^{∘} = H_{i,α}/H_{i,α} ∩ P is a projective homogeneous variety.

We fix a simple basis of Φ_{α}^{i}(x, A) such that V_{i}(x, A) is spanned by simple roots. We may moreover assume that α is the simple root α_{j} that is not contained in V_{i}(x, A). Then y is stabilised by the one dimensional torus S_{j} that is the connected component of < s ∈ S|χ(s) = 1, ∀χ ∈ V_{i}(x, A) > that contains the identity. By assumption y is also stabilised by the unipotent element h. Let F be the group generated by h and S_{j}(K). Then F is a subgroup of H_{i,α}(K).

We now consider I(y) inside the building of H_{i,α}(K). Inside the buiding of H_{i,α}(K) one has that I_{A}^{i}(y) is the line S_{j}(𝕂) · z for some suitable point z ∈ I_{A}^{i}(y). Since F stabilises the point y, it follows that I_{fA}^{i}(y) = f(I_{A}^{i}(y)) for all elements f ∈ F. In particular the tree F (I_{A}^{i}(y)) is a subset of I(y), since ρ_{A}(y) = ρ(y).

Let us fix a Borel subgroup B of H_{i,α} that stabilises a sector that has a halfline in common with I_{A}^{i}(x). We can choose B in such a way that H_{α}^{i}(x, A) is contained in B. The opposite Borel group B^{−} also stabilises a sector that has a halfline in common with I_{A}^{i}(y). Moreover B^{−} contains the group H_{−α}^{i}(x, A).

Let f ∈ B ∩ F be such that I_{fA}^{i}(y) ∩ I_{A}^{i}(y) is a halfline. We can find a unipotent element g ∈ B^{−} such that gA and fA have a sector in common. In fact we can choose g in such a way that f(I_{A}^{i}(y)) ∩ gA contains a halfline (which is the complement in I_{fA}^{i}(y) of I_{fA}^{i}(y) ∩ I_{A}^{i}(y)).

Since g ∈ B^{−} we can write g uniquely as g = g_{1} · g_{2} with g_{1} ∈ H_{−α}^{i}(x, A) and g_{2} = ∏ u_{γ}(b_{γ}) with γ ∈ Φ_{α}^{i}(x, A)^{−} − Ψ_{−α}^{i}(x, A). We will now show that g_{1} stabilises y.

Since the roots γ occurring in the expression for g_{2} above are all contained in V_{i}(x, A), it follows that g_{2}(I_{A}^{i}(y)) = I_{A}^{i}(y). Hence I_{g_{1}A}(y) contains the line g(I_{A}^{i}(y)). Using the commutator relations we will show that this implies that g_{1} stabilises the point y.

Let g_{1} be g_{1} = u_{δ_{m}}(d_{m}) ⋯ u_{δ_{1}}(d_{1}) with δ_{j} ∈ Ψ_{−α}^{i}(x, A) for j = 1, …, m. We may assume that y has the form y = ∏ u_{β}(b_{β}) · P with β ∈ Φ ∩ V_{i}(x, A).

We first move u_{δ_{1}}(d_{1}) using the commutator relations. If (δ_{1}, λ) < 0 we do not move u_{δ_{1}}(d_{1}). If (δ_{1}, λ) ≥ 0, then we move the term u_{δ_{1}}(d_{1}) to the right until it disappears into P. Now some terms u_{β+δ_{1}} might have appeared because of the commutator relations. If (β + δ_{1}, λ) < 0 then we move this term to the left until we arrive at the term u_{δ_{2}}(d_{2}). If (β + δ_{1}, λ) ≥ 0, then we move this term to the right until it disappears into P. After a finite number of steps this proces stops. Now we do the same with the term u_{δ_{2}}(d_{2}).

After a finite number of steps we are finished. Now we find that g_{1}(y) = ∏ u_{η}(c_{η})·y, where the roots η are such that (η, λ) < 0 and moreover η ∈ Ψ_{−α}^{i}(x, A). Since g(I^{i}A(y)) ⊂ I_{g_{1}A}(y), it follows that for all η one must have c_{η} = 0. Hence g_{1}(y) = y.

The group F^{−} generated by g_{1} and S_{j}(K) stabilises y. One observes that the group generated by F and F^{−} acts transitively on the tree F (I_{A}^{i}(y)). In fact the union of all the lines I_{A′}^{i}(y) for apartments A′ with ρ_{A′}(y) = ρ(y) gives us a tree on which the stabiliser H̃ of y acts transitively. Now the groups < S_{j}(K), H_{α}^{i}(x, A) ∩ H̃ > and < S_{j}(K), H_{−α}^{i}(x, A) ∩ H̃ > generate a semisimple group H that acts transitively on this tree. In fact these two groups form a pair of opposite Borel groups inside H. It follows from the construction that this group has K-rank one. 8.9. Remark. It follows from the fact that all elements of the group H stabilise the point y, that for any element g of H ∩ H_{α}^{i}(x, A) ∩ P_{z} one has: g ∉ P (I_{A}^{i}(x)) ⟹ I_{gA}^{i}(x) ≠ I_{A}^{i}(x) ∩ gA. Moreover the same statement is also true for elements g ∈ H ∩ H_{−α}^{i}(x, A) ∩ P_{z}. 8.10. Remark. We can now apply the classification of conjugacy classes of unipotent elements in semisimple groups as described in [C] Ch. 5 and §13.1 to our situation. In particular we can use the theory of weighted Dynkin diagrams. Even though the group H in the proposition above might not be P SL_{2}(K) or SL_{2}(K), since it might be a non split form of some other group. However it always contains subgroups SL_{2}(K) or P SL_{2}(K). Therefore we may assume, after replacing h by another unipotent element of H if necessary, that h is contained in a group SL_{2}(K).

Let Ψ be the irreducible component of the root system Φ_{α}^{i}(x, A) that contains the root α. Let us fix a simple basis of Ψ such that Ψ ∩ V_{i}(x, A) is spanned by simple roots. Then we may moreover assume that α is the simple root α_{j} that is not contained in V_{i}(x, A). It now follows from the construction of H in the proof of proposition 8.8 above, that the maximal K−split torus of H (and hence of our group SL_{2}(K) containing h) is the 1-ps of S that corresponds to a multiple of the fundamental weight ω_{j}. Hence the weighted Dynkin diagram belonging to our unipotent element h is the Dynkin diagram of Ψ with all weights equal to zero except one. The unique non-zero weight corresponds to the vertex that represents the simple root α_{j}.

We now give some examples of the group H that stabilises a point y as constructed in proposition 8.8. 8.11. Example. Let Char(K) ≠ 2. Let us take a basis e_{1}, …, e_{4}, f_{1}, …, f_{4}. of K^{8}. On K^{8} we define an inner product g by g(e_{i}, f_{j}) = δ_{ij} and g(f_{i}, f_{j}) = g(e_{i}, e_{j}) = 0. Let G := SO(g). Then Φ = D_{4}. Let X consist of the points x ∈ ℙ^{7} such that g(x, x) = 0. Even though the group G is not simply connected our construction still applies in this case. Let us choose coordinates such that a general point x is of the form x = x_{1} · e_{1} + ⋯ + x_{4} · e_{4} + y_{1} · f_{1} + ⋯ + y_{4} · f_{4}.

We take a point y such that x_{4} = y_{4} = 0, x_{2} = y_{2} = a, x_{3} = y_{3} = b, x_{1} = c, y_{1} = d. Here a, b, c, d ∈ 𝕂 and all have absolute value equal to one. We moreover assume that there are no \overline{K}-rational linear relations between \overline{a}, \overline{b}, \overline{c} and \overline{d}. Here \overline{K} denotes the residue field of K and \overline{a}, \overline{b}, \overline{c} and \overline{d} are the images of a, b, c and d in \overline{K} = K^{∘}/πK^{∘}. Let A ⊂ B be the apartment corresponding to the torus in G that acts diagonally w.r.t. our basis.

We will now determine the stabiliser of y in G(K). The stabiliser of y also stabilises the variety of the K-rational points in the variety y^{⊥} ∩ X. We determine these points. From g(z, y) = 0 it follows that dx_{1} +cy_{1} +a(x_{2} +y_{2})+b(x_{3} +y_{3}) = 0. Since we assume z to be a K valued point and there are no K- rational relations between a, b, c and d, it follows that one has that x_{1} = y_{1} = 0, x_{2} + y_{2} = 0 and x_{3} + y_{3} = 0. So z = z_{2}(e_{2} − f_{2}) + z_{3}(e_{3} − f_{3}) + z_{4}e_{4} + z_{5}f_{5}. Since g(z, z) = 0, the coordinates z_{i} satisfy the relation −z_{2}^{2} − z_{3}^{2} + z_{5}z_{4} = 0. Therefore the stabiliser H of y is equal to the automorphism group of this quadratic form. The K-rank of H is one if and only if this form has Witt index one over K. In that case it follows that ρ_{A}(y) = ρ(y). Otherwise ρ_{A}(y) > ρ(y).

8.12. Example. Let Φ = A_{ℓ} consist of the roots e_{i} − e_{j} with i ≠ j and 1 ≤ i, j ≤ ℓ + 1. Let α_{i} := e_{i} − e_{i+1}, i = 1, …, ℓ be a simple basis of Φ. We assume that ℓ = 2s + 1 for some integer s ≥ 1. Let λ be the weight λ = ∑_{n=0}^{s} α_{2n+1}. Let P ⊂ G := SL_{ℓ+1} be the parabolic subgroup determined by λ. Let X := G/P. Let x := ∏_{n=0}^{s} u_{−α_{2n+1}}(x_{n}) · P ∈ X. We assume that |x_{n}| = 1 for n = 0, …, s. Moreover we require that ax_{n} +b ≠ 0 mod π for all a, b ∈ (K^{∘})^{∗} for all n. Then V := V_{1}(x, A) is spanned by the roots α_{2n+1}, n = 0, …, s. The conditions also ensure us that ρ_{A}(y) = ρ(y) holds.

Let us also assume that x_{n}/x_{m} ∈ K^{∗} for 0 ≤ n, m ≤ s. We will now describe the stabiliser of y in SL_{ℓ+1}(K). Let h_{n,m}(a, b) for n ≠ m and 0 ≤ n, m ≤ s be the element u_{β}(a) · u_{γ}(b), where β = e_{2n+1} − e_{2m+1} and γ = e_{2n+2} − e_{2m+2}. Note that γ ≡ β mod V. Using the commutator relations one verifies that h_{n,m}(a, b) · y = y if and only if d_{1}ax_{n} + d_{2}bx_{m} = 0. Here d_{1} (resp. d_{2}) is the structure constant occurring in the commutator relations for β and −α_{2n+1} (resp. γ and −α_{2m+1}). We may assume that they are ±1.

Let H_{n,m}, n ≠ m be the group generated by the elements h_{n,m}(a, b) that stabilise y. Our conditions ensure us that H_{n,m} ≠ {1}. In fact < H_{n,m}, H_{m,n} > is a group isomorphic to SL_{2}(K). The stabiliser H of y is the group generated by the groups H_{n,m}, n ≠ m, 0 ≤ n, m ≤ s. One verifies that H = SL_{s+1}(K). Note that the root system A_{s} of the group H is the quotient root system Φ mod Φ ∩ V in the sense of [Ho] theorem 6 or [C] §10.4.

8.13. Definition. Let (Φ, K, Y^{ss}) not be disturbed. Let x ∈ Y^{s}. We say that I(x) solely depends on the absolute value of the coordinates if for all apartments A that determine I(x) either m(x, A) = 1 or m(x, A) > 1 and for all 1 ≤ i < m(x, A) one has for all z ∈ I_{A}^{i}(x) and for all roots α ∈ Φ such that α ∉ V_{i}(x, A) and α ⊥̸ V_{i}(x, A), that for all h ∈ H_{α}^{i}(x, A)^{∗} ∩ P_{z} the following holds: h ∉ P (I_{A}^{i}(x)) ⟹ I_{hA}^{i}(x) = I_{A}^{i}(x) ∩ hA.

Here H_{α}^{i}(x, A)^{∗} = H_{α}^{i}(x, A) if Ψ_{α}^{i}(x, A) does not contain a root γ ⊥⊥_{x} V_{i}(x, A). If Ψ_{α}^{i}(x, A) does contain a root γ ⊥⊥_{x} V_{i}(x, A), then H_{α}^{i}(x, A)^{∗} := ∏ U_{β}, where the roots β are in Ψ_{α}^{i}(x, A) − {γ}. Here we have choosen a fiixed order of the roots. Note that the set H_{α}^{i}(x, A)^{∗} depends on the order of the roots choosen. However the property defined above is independent from the order choosen.

Since (Φ, K, Y^{ss}) is not disturbed, it follows that if γ ⊥⊥_{x} V_{i}(x, A) then γ ⊥⊥ V_{i}(x, A). Let us assume that Ψ_{α}^{i}(x, A) contains a root γ ⊥⊥ V_{i}(x, A). Then the group H constructed in proposition propostion 8.8 as the stabiliser of a point y, always contains a group SL_{2}(K) that is generated by the groups U_{γ}(K) and U_{−γ}(K). The condition above now states that in fact H is equal to this subgroup SL_{2}(K).

We will now give some examples of spaces Y^{s} such that I(x) solely depends on the absolute value of the coordinates for all x ∈ Y^{s}.

8.14. Example. Let X = G/B and let Y^{ss} ⊂ X(𝕂). The stabiliser of a point y ∈ Y^{ss} is contained in a Borel group and cannot contain a non-trivial semisimple subgroup. Therefore I(x) is solely determined by the absolute value of the coordinates for all points x ∈ Y^{s}.

Since λ ∈ V_{1}(x, A), it follows that there are no roots that are orthogonal to V_{1}(x, A) for all x ∈ Y^{ss} and for all apartments A ⊂ B. Therefore the group H(x, A) defined in 7.13 is the group P (𝕀(x)). The theorem below now implies that I(x) = I_{A}(x) for some apartment A ⊂ B. (This remains true for all root systems.)

To be able to give some less trivial examples we collect some results from the literature. They concern the case where the group H_{α}^{i}(x, A) is abelian.

8.15. Proposition. Let x ∈ Y_{A}^{ss} be such that m(x, A) > 1. Let 1 ≤ i < m(x, A). Let H_{i,α} :=< U_{β} | β ∈ Φ_{α}^{i}(x, A) > and let L_{i,α} ⊂ H_{i,α} be the Levi subgroup of the maximal parabolic subgroup P_{i,α} that has H_{α}^{i}(x, A) as its unipotent radical. We assume that α is such that for all β ∈ Ψ_{α}^{i}(x, A) one has that β ≡ α mod V_{i}(x, A). If Φ_{α}^{i}(x, A) has roots of different lengths then we moreover assume that Char(K) ≠ 2. Let h ∈ H_{α}^{i}(x, A).

Then there exists an element g ∈ L_{i,α}(K) such that g · h · g^{−1} = ∏ u_{δ_{j}}(a_{j}), where the roots δ_{j} ∈ Ψ_{α}^{i}(x, A) are mutually orthogonal and are all long roots.

Proof. This has been proved in [RRS] proposition 2.13 for algebraically closed fields. One verifies that the proof remains valid if the field is not algebraically closed. In particular that one then can take g in L_{i,α}(K).

8.16. Corollary. Let us keep the notation as in proposition 8.15 above. Let z ∈ I_{A}^{i}(x). Let h ∈ H_{α}^{i}(x, A) be such that h ∉ P (I_{A}^{i}(x)).

Then I_{hA}^{i}(x) ≠ I_{A}^{i}(x) ∩ hA can only occur if ∑ δ_{j} ⊥ V_{i}(x, A).

Proof. Let us assume that I_{hA}^{i}(x) ≠ I_{A}^{i}(x) ∩ hA. We may moreover assume that the element h is brought into a standard form such that h = ∏ u_{δ_{j}}(a_{j}) with the roots δ_{j} mutually orthogonal and long. Let us now consider the point y constructed in 8.7, that is stabilised by h

We can find a group SL_{2}(K) containing h using the same variety Z^{∘} as in proposition 8.8, but taking the stabiliser of y inside the group (SL_{2})^{s} =< U_{δ_{j}}, U_{−δ_{j}}| j = 1, …, s > similarly to the proof of proposition 8.8. We get a group H = SL_{2}(K) that is diagonally embedded in the group (SL_{2})^{s}. In particular its maximal torus corresponds to the weight ∑ 2δ_{j}/(δ_{j}, δ_{j}). If this weight is not a multiple of the weight ω_{j} (See the proof of proposition 8.8.), then the stabiliser of y (in the group H_{i,α}) would contain a torus of K-rank two. This cannot be, since ρ_{A}(y) = ρ(y). Since all the roots δ_{j} have the same length, the corollary follows.

8.17. Remark. Le W be the Weyl group of Φ_{α}^{i}(x, A). Let w ∈ W be the product of the reflections w_{δ_{j}} in the hyperplanes orthogonal to the roots δ_{j}, j = 1, …, s (See 8.15). Let V := V_{i}(x, A).

Let us assume that ∑ δ_{j} ⊥ V. Then w preserves V. Let ω_{j} be the fundamental weight orthogonal to V. Then w(ω_{j}) = −ω_{j}.

We will now show that w in fact preserves a simple basis of Φ_{α}^{i}(x, A) ∩ V. There exists a simple basis of Φ_{α}^{i}(x, A), such that the roots δ_{j}, j = 1, …, s, are simple. For all roots β ∈ Φ_{α}^{i}(x, A) ∩ V one has that β is orthogonal to either all roots δ_{j} or to all but two roots δ_{j}. In the latter case, it follows from ∑ δ_{j} ⊥ V that the signs must be opposite. From this it follows that ht(w(β)) = ht(β). Here the height is taken w.r.t our simple basis. Therefore w preserves the simple basis of Φ_{α}^{i}(x, A)∩V that is determined by the set of positive roots contained in V that is determined by our simple basis of Φ_{α}^{i}(x, A) that contains the roots δ_{j}.

Therefore we can take a simple basis of Φ_{α}^{i}(x, A) such that V is spanned by simple roots and moreover the set of positive roots that are contained in V is preserved by w. Let w_{2} be the longest element of the Weyl group of Φ_{α}^{i}(x, A) and w_{1} be the longest element of the Weyl group of Φ_{α}^{i}(x, A) ∩ V. Then w = w_{2}w_{1}.

8.18. Examples. We give some examples of weights λ such that the corresponding flag domain Y^{s} is such that I(x) is solely determined by the absolute value of the coordinates. Let x ∈ Y_{A}^{s} with m(x, A) > 1. Let 1 ≤ i < m(x, A) and ρ_{A}(x) = ρ(x). Let α ∈ Φ be such that α ⊥̸ V := V_{i}(x, A) and α ∉ V. Let Ψ be the irreducible component of Φ_{α}^{i}(x, A) that contains the root α.

We first consider the root system Φ = A_{ℓ}. Then H_{α}^{i}(x, A) is always abelian. Moreover Ψ∩V consists of at most two irreducible components. The element w_{1}w_{2} (See 8.17) preserves Ψ ∩ V if and only if Ψ ∩ V = A_{j} × A_{j} and Ψ = A_{2j+1} (See [Ho] lemma 10).

Therefore if the weight λ is such that for all x ∈ Y^{s} one has that Ψ ∩ V never has two irreducible components A_{j}, then I(x) is solely determined by the absolute value of the coordinates for all x ∈ Y^{s}. An example of such a weight is λ = ω_{1} +ω_{ℓ}. Then λ is a root. It now follows from proposition 5.4 that Ψ∩V is alway irreducible, that is of the form A_{j}.

Next we give an example for the root system Φ = C_{ℓ}. Let Φ consist of the roots ±2e_{i}, ±e_{i} ± e_{j} with i ≤ i, j ≤ ℓ and i ≠ j. We choose λ = e_{1}. Then X = ℙ^{2ℓ−1}. It follows from proposition 5.4 that Ψ ∩ V is irreducible of type C_{j}. Here C_{1} = A_{1} is consists of a long root. So we have only one root α to consider in this case.

Now Ψ_{α}^{i}(x, A) always contains a long root γ ⊥⊥ V. By conjugating h ∈ H_{A}^{i}(x, A) with an element of L_{i,α} (see 8.15), one can write ghg^{−1} = u_{δ}(a)·u_{γ}(b). (See also the remark in the introduction of [Rö]). Such an element can only stabilise a point y (as in proposition 8.7) if a = 0. Therefore I(x) is solely determined by the absolute value of the coordinates for all x ∈ Y^{s}. 8.19. Definition. Let x ∈ Y_{A}^{ss} and z ∈ I_{A}^{m(x,A)}(x). Let B be a Borel group and U ⊂ B be its maximal unipotent subgroup. Let h ∈ U ∩ P_{z}. We choose an ordering on the set of positive roots determined by B. Let Φ^{+} consist of the roots α_{1}, α_{2}, . . ..

For α_{j} ∈ Φ^{+} we define H_{j} as being the smallest half space bounded by a wall of type α_{j} that contains A ∩ h · A. If there exists an integer i, 1 ≤ i < m(x, A), such that α_{j} ⊥ V_{i}(x, A) (and α_{j} ⊥⊥_{x} V_{i−1}(x, A) if i > 1), but not α_{j} ⊥⊥_{x} V_{i}(x, A), then we put H_{j}^{†} := s · H_{j}. Here s ∈ S(𝕂) is as in proposition 7.11. In all other cases we take H_{j}^{†} := H_{j}. Note that if (Φ, K, Y^{ss}) is not disturbed then H_{j}^{†} = H_{j} for all roots α_{j}.

Let j_{0} := 0 and let j_{i} := ♯Φ^{+} − ♯{α ∈ Φ^{+} | α ⊥⊥_{x} V_{i}(x, A)} if 1 ≤ i ≤ m(x, A). We say that our ordering of the positive roots is fair dinkum for x, A and h if the following two statements hold:

i) α_{n} ⊥⊥_{x} V_{i}(x, A) if and only if j_{i} < n.

ii) If j_{i} < n < m ≤ j_{i+1} then either 𝕀_{A}^{i}(x) ∩ ⋂_{j=j_{i}}^{n} H_{j}^{†} is a proper subset of

𝕀_{A}^{i}(x) ∩ ⋂_{j=j_{i}}^{m} H_{j}^{†} or 𝕀_{A}^{i}(x) ∩ H_{n}^{†} = 𝕀_{A}^{i}(x) ∩ H_{m}^{†}.

8.20 Lemma. Let A_{1}, A_{2} ⊂ B be two apartments. Let z_{i} ∈ A_{i}, i = 1, 2. Then there exists an apartment A such that A ∩ A_{i}, i = 1, 2 contains a sector S_{i} with base point z_{i}. In A the sectors S_{1} and S_{2} are opposite.

In particular A = h_{i}A_{i} with h_{i} ∈ P_{z_{i}} ∩ B_{i}. Here B_{i} is the Borel group associated to S_{i}. Proof. This is well known. It is a fairly straightforward consequence of the theorem in [B] Ch. VI §8. 8.21 Theorem. Let (Φ, K, Y^{ss}) not be disturbed. Let Y^{s} be such that I(x) is solely determined by the absolute value of the coordinates for all x ∈ Y^{s}. Let x ∈ Y^{s} and A_{1}, A_{2} ⊂ B be apartments that determine I(x). Then A_{2} = h · A_{1} with h ∈ H(x, A_{1}). Proof. Let z_{j} := I_{Aj}^{m(x,A_{j})}(x) for j = 1, 2. Let A be an apartment containing z_{1} and z_{2} and such that A ∩ A_{j}, j = 1, 2 contains a sector. Then A = gA_{1} = fA_{2}, where g (resp. f) is an element of the Borel subgroup B_{1} (resp. B_{2}) of G stabilising the sector in A ∩ A_{1} (resp. A ∩ A_{2}).

Let Φ^{+} be the set of positive roots, determined by the Borel subgroup B_{1}. We choose an ordering of the positive roots, that is fair dinkum for x, A_{1} and g. Now we can write g as g = u_{α_{j_{m(x,A_{1})}}} ⋯ u_{α_{1}}(a_{1}). We let g_{i} be g_{i} := u_{α_{j_{i}}} (a_{j_{i}}) ⋯ u_{α_{j_{i−1}+1}}(a_{j_{i−1}+1}). Then g = g_{m(x,A)} ⋯ g_{1}. For f we make a similar decomposition f = f_{m(x,A_{2})} ⋯ f_{1}.

If g_{i} ∈ H_{i}(x, A_{1}), i = 1, …, m(x, A_{1}), then g ∈ H(x, A_{1}). Then g(𝕀_{A_{1}}(x)) ⊂ I_{A}(x). Since ρ_{A}(x) = ρ(x), one has g(𝕀_{A_{1}}(x)) = 𝕀_{A}(x). In particular we have I_{A_{1}}^{m(x,A_{1})}(x) = I_{A}^{m(x,A)}(x). From ρ_{A}(x) = ρ_{A_{2}}(x) = ρ(x), one concludes that A = fA_{2} with f ∈ H(x, A_{2}) = H(x, A) = H(x, A_{1}). From this it follows that A_{2} = hA_{1} with h ∈ H(x, A_{1}).

So let us assume that g ∉ H(x, A_{1}). Let n ≥ 1 be the smallest integer such that g_{n} ∉ H_{n}(x, A_{1}). If n > 1, then we can replace A_{1} by g_{n−1} ⋯ g_{1}A_{1} without changing 𝕀_{A_{1}}(x). Moreover, if n > 1 then also f_{n−1}, …, f_{1} ∈ H(x, A_{2}). Therefore we can also replace A_{2} by f_{n−1} ⋯ f_{1} · A_{2}.

Since g_{n} ∉ H_{n}(x, A_{1}), one has I_{g_{n}A_{1}}^{n}(x) ⊂ 𝕀_{A_{1}}^{n}(x) ∩ g_{n}A_{1}. This follows form the fact that I(x) is solely determined by the absolute value of the coordinates. We now consider the faces of I_{g_{n}A_{1}}^{n}(x) that are in the interior of 𝕀_{A_{1}}^{n}(x). The roots that correspond to these faces (See definition 5.5) are positive for the ordering determined by the Borel subgroup B_{1}. These faces will not be changed by the elements g_{j} with j > n, since these g_{j} are products of elements of the groups U_{α} with α ⊥⊥_{x} V_{n}(x, A). However they can add extra faces to I_{A}^{n}(x) that are also in the interior of 𝕀_{A_{1}}^{n}(x). Again these faces correspond to roots that are positive for the Borel group B_{1}. Therefore the faces of I_{A}^{n}(x) that are in the interior of 𝕀_{A_{1}}^{n}(x) are all positive for the ordering determined by B_{1}.

We can now do the same for A_{2} and f_{n}. We now we consider the faces of I_{A}^{n}(x) that are in the interior of 𝕀_{A_{2}}^{n}(x). The roots corresponding to these faces are all positive for the Borel group B_{2}. Hence they are negative for B_{1}. This cannot be.

Hence we must have g_{n} ∈ H_{n}(x, A_{1}) and f_{n} ∈ H_{n}(x, A_{2}). Using induction we find that g ∈ H(x, A_{1}) and f ∈ H(x, A_{2}). Therefore A_{2} = hA_{1} with h ∈ H(x, A_{1}) = H(x, A) = H(x, A_{2}).

8.22. Remark. Let us keep the assumptions of theorem 8.21. Then it follows from the theorem that for all x ∈ Y^{s} the set {I_{A}^{m(x,A)}(x)| ρ_{A}(x) = ρ(x)} consists of a single point. So one can use this to define a G(K)- equivariant map from Y^{s} to the building B. One can not expect that this map coincides with the map defined in 3.17, using the circumcenter of I(x).

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