Spherical Functions on a Group of $p\text{-adic}$ Type

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 February 2014

Chapter II: Groups of $p\text{-adic}$ type

Root systems

Let $A$ be a real vector space of finite dimension $l>0,$ and $⟨x,y⟩$ a positive definite scalar product on $A\text{.}$ Let $A^{*}$ be the dual vector space. Then the scalar product defines an isomorphism of $A$ onto $A^{*},$ and hence a scalar product $⟨a,b⟩$ on $A^{*}\text{.}$ For each non-zero $a\in A^{*}$ let $a^{\vee }\in A$ be the image of $2a/⟨a,a⟩$ under this isomorphism. Let $h_a$ denote the kernel of $a,$ and let $w_a$ be the reflection in the hyperplane $h_a\text{:}$ $$w_a\left(x\right)=x-a\left(x\right)a^{\vee }(x\in A)\text{.}$$ We make $w_a$ act also on $A^{*},$ by transposition: $w_a\left(b\right)=b\circ w_a\text{.}$

A root system ${\Sigma }_0$ in $A^{*}$ is a subset ${\Sigma }_0$ of $A^{*}$ satisfying the following axioms:

(RS 1)	${\Sigma }_0$ is finite, spans $A^{*}$ and does not contain $0\text{;}$
(RS 2)	$a\in {\Sigma }_0\Rightarrow w_a\left({\Sigma }_0\right)={\Sigma }_0\text{;}$
(RS 3)	$a,b\in {\Sigma }_0\Rightarrow a\left(b^{\vee }\right)\in \mathbb{Z}\text{.}$

We shall assume throughout that ${\Sigma }_0$ is reduced:

(RS 4)

If $a,b\in {\Sigma }_0$ are proportional, then $b=\pm a\text{;}$

and irreducible:

(RS 5)

If ${\Sigma }_0={\Sigma }_1\cup {\Sigma }_2$ such that $⟨a_1,a_2⟩=0$ for all $a_1\in {\Sigma }_1$ and $a_2\in {\Sigma }_2,$ then either ${\Sigma }_1$ or ${\Sigma }_2$ is empty.

The elements of ${\Sigma }_0$ are called roots. The group $W_0$ generated by the reflections $w_a$ $(a\in {\Sigma }_0)$ is a finite group of isometries of $A,$ called the Weyl group of ${\Sigma }_0\text{.}$ The connected components of the set $A-\bigcup _{a\in {\Sigma }_0}{h}_a$ are open simplicial cones called the chambers of ${\Sigma }_0\text{.}$ These chambers are permuted simply transitively by $W_0,$ and the closure of each chamber is a fundamental region for $W_0\text{.}$

Choose a chamber $C_0\text{.}$ This choice determines a set ${\Pi }_0$ of $l$ roots such that $C_0$ is the intersection of the open half-spaces $\{x\in A:a\left(x\right)>0\}$ for $a\in {\Pi }_0\text{.}$ The elements of ${\Pi }_0$ are called the simple roots (relative to the chamber $C_0\text{).}$ A root $b$ is said to be positive or negative according as it is positive or negative on $C_0\text{.}$ Let ${\Sigma }_0^{+}$ (resp. ${\Sigma }_0^{-}\text{)}$ denote the set of positive (resp. negative) roots.

Each root $b$ can be expressed as a linear combination of the simple roots: $$b=\sum _{a\in {\Pi }_0}{m}_{a}a$$ with coefficients $m_a\in \mathbb{Z}$ all of the same sign. The integer $\sum m_a,$ the sum of the coefficients, is called the height of $b\text{.}$ There is a unique root (the 'highest root') whose height is maximal.

Let $C_0^{\perp }$ denote the 'obtuse cone' $$C_0^{\perp }=\{x\in A:⟨x,y⟩\ge 0\hspace{0.17em}\text{for all}\hspace{0.17em}y\in {\bar{C}}_0\}\text{.}$$ If $x\in {\bar{C}}_0$ and $w\in W_0,$ then $x-wx\in C_0^{\perp }\text{.}$

A subset ${\Phi }_0$ of ${\Sigma }_0$ is said to be closed if $$a,b\in {\Phi }_0\hspace{0.17em}\text{and}\hspace{0.17em}a+b\in {\Sigma }_0\Rightarrow a+b\in {\Phi }_0\text{.}$$ Suppose ${\Phi }_0$ is closed and ${\Phi }_0\cap -{\Phi }_0=\varnothing \text{.}$ Then there exists $w\in W_0$ such that $w{\Phi }_0\subseteq {\Sigma }_0^{+}\text{.}$

The Weyl group $W_0$ is generated by the reflections $w_a$ for $a\in {\Pi }_0\text{.}$ Hence any $w\in W$ can be written as a product $w=w_1\cdots w_r,$ where $w_i=w_{a_i}$ and $a_i\in {\Pi }_0$ $(1\le i\le r)\text{.}$ If the number of factors $w_i$ is as small as possible, then $w_1\cdots w_r$ is a reduced word for $w,$ and $r$ is the length of $w,$ written $l\left(w\right)\text{.}$ The length of $w$ is also equal to the number of positive roots $b$ such that $w^{-1}b$ is negative. More precisely, these roots are $b_i=w_1\cdots w_{i-1}{a}_i$ $(1\le i\le r)\text{.}$ (All this of course is relative to the choice of the chamber $C_0\text{.)}$

In particular, there exists a unique element $w_0$ of $W_0$ which sends every positive root to a negative root. This leads to the following result, which will be useful later:

The positive roots can be arranged in a sequence $(b_1,\dots ,b_n)$ such that for each $r=0,1,\dots ,n$ the set of roots $$\{-b_1,-b_2,\dots ,-b_r,b_{r+1},b_{r+2},\dots ,b_n\}$$ is the set of positive roots relative to some chamber $C_r\text{.}$

		Proof.
	Let $w_1{w}_2\cdots w_n$ be a reduced word for $w_0,$ where $w_i=w_{a_i},$ $a_i\in {\Pi }_0,$ and as before put $b_i=w_1{w}_2\cdots w_{i-1}{a}_i$ $(1\le i\le n)\text{.}$ For each $r,$ $w_{r+1}\cdots w_n$ is a reduced word, and therefore from the facts quoted above $$\{a_{r+1},w_{r+1}{a}_{r+2},\dots ,w_{r+1}{w}_{r+2}\dots w_{n-1}{a}_n\}={\Sigma }_0^{+}\cap \left(w_{r+1}\dots w_n{\Sigma }_0^{-}\right)\text{.}$$ Consequently $$\begin{array}{ccc}\{b_{r+1},b_{r+2},\dots ,b_n\}& =& \left(w_1\cdots w_r{\Sigma }_0^{+}\right)\cap \left(w_0{\Sigma }_0^{-}\right)\\ & =& \left(w_1\cdots w_r{\Sigma }_0^{+}\right)\cap {\Sigma }_0^{+}\end{array}$$ and therefore $$\{b_1,b_2,\dots ,b_r\}=\left(w_1\cdots w_r{\Sigma }_0^{-}\right)\cap {\Sigma }_0^{+}$$ so that $$\{-b_1,-b_2,\dots ,-b_r\}=\left(w_1\cdots w_r{\Sigma }_0^{+}\right)\cap {\Sigma }_0^{-}$$ and finally $$\{-b_1,\dots ,-b_r,b_{r+1},\dots ,b_n\}=w_1\cdots w_r{\Sigma }_0^{+}$$ is the set of positive roots for the chamber $C_r=w_1\cdots w_r{C}_0\text{.}$ $\square$

Affine roots

We retain the notation of (2.1). For each $a\in {\Sigma }_0$ and $k\in \mathbb{Z}$ we have an affine-linear function $a+k$ on $A$ (namely, $x\mapsto a\left(x\right)+k\text{).}$ Let $$\Sigma =\{a+k:a\in {\Sigma }_0,k\in \mathbb{Z}\}\text{.}$$ The elements of $\Sigma$ are called affine roots. We shall denote them by Greek letters $\alpha ,\beta ,\dots \text{.}$ If $\alpha$ is an affine root, so are $-\alpha$ and $\alpha +k$ for all integers $k\text{.}$ If $\alpha =a+k$ then $a$ $(\in {\Sigma }_0)$ is the gradient of $\alpha$ (in the usual sense of elementary calculus).

For each $\alpha \in \Sigma ,$ let $h_{\alpha }$ be the (affine) hyperplane on which $\alpha$ vanishes, and let $w_{\alpha }$ be the reflection in $h_{\alpha }\text{.}$ (From now on, only the affine structure and not the vector-space structure of $A$ will come into play). The group $W$ generated by the $w_{\alpha }$ as $\alpha$ runs through $\Sigma$ is an infinite group of displacements of $A,$ called the Weyl group of $\Sigma$ or the affine Weyl group. $W_0$ is the subgroup of $W$ which fixes the point $0\text{.}$ The translations belonging to $W$ form a free abelian group $T$ of rank $l,$ and $W$ is the semi-direct product of $T$ and $W_0\text{.}$

For each $a\in {\Sigma }_0$ let $$t_a=w_a\circ w_{a+1}\in T\text{.}$$ The $t_a$ $(a\in {\Pi }_0)$ are a basis of $T\text{.}$ We have $t_a\left(0\right)=a^{\vee },$ and the mapping $T\to A$ defined by $t\mapsto t\left(0\right)$ maps $T$ isomorphically onto the lattice spanned by the $a^{\vee }$ for $a\in {\Sigma }_0\text{.}$

The affine Weyl group acts on $\Sigma$ by transposition: if $w\in W,$ $\alpha \in \Sigma$ then $w\left(\alpha \right)$ is defined to be $\alpha \circ w^{-1}\text{.}$

The connected components of the set $A-\bigcup _{\alpha \in \Sigma }{h}_{\alpha }$ are open rectilinear $l\text{-simplexes}$ called the chambers of $\Sigma \text{.}$ The chambers are permuted simply transitively by $W,$ and the closure of each chamber is a fundamental region for $W\text{.}$ The (non-empty) faces of the chambers are locally closed simplexes, called facets. Each point $x\in A$ lies in a unique facet $F\text{.}$ The subgroup of $W$ which fixes $x$ also fixes each point of $F\text{.}$

Choose a chamber $C\text{.}$ This choice determines a set $\Pi$ of $l+1$ affine roots, such that $C$ is the intersection of the open half-spaces $\{x\in A:\alpha \left(x\right)>0\}$ for $\alpha \in \Pi \text{.}$ The affine roots belonging to $\Pi$ are called the simple affine roots (relative to the chamber $C\text{).}$

The affine Weyl group $W$ is generated by the relations $w_{\alpha }$ for $\alpha \in \Pi \text{.}$ As in the case of $W_0,$ we define the length $l\left(w\right)$ of an element $w$ of $W\text{.}$ The length of $w$ is also equal to the number of affine roots $\alpha$ which are positive on $C$ and negative on $wC,$ i.e. to the number of hyperplanes $h_{\alpha }$ which separate $C$ and $wC\text{.}$

If $F$ is a facet of $C,$ then the subgroup of $W$ which fixes a point of $F$ is generated by the $w_{\alpha }$ $(\alpha \in \Pi )$ which fix $x\text{.}$

We shall need the following result later:

Let $\Phi$ be a set of affine roots whose gradients are all positive and distinct. Then there exists a translation $t\in T$ such that each $\alpha \in \Phi$ is positive on $t\left(C_0\right)\text{.}$

		Proof.
	Let the elements of $\Phi$ be $a+r_a$ $(a\in {\Sigma }_0^{+})$ and let $t\in T\text{.}$ We have $$(a+r_a)\left(t\left(x\right)\right)=a(x+t\left(0\right))+r_a=a\left(x\right)+a\left(t\left(0\right)\right)+r_a\text{.}$$ If $x\in C_0$ then $a\left(x\right)>0$ for all positive roots $a\text{.}$ Hence $a+r_a$ will be positive on $t\left(C_0\right)$ provided that $a\left(t\left(0\right)\right)+r_a\ge 0\text{.}$ Hence we have only to choose $t$ so that $t\left(0\right)$ satisfies $$a\left(t\left(0\right)\right)\ge -r_a$$ for all $a\in {\Sigma }_0^{+}$ which occur as gradients of elements of $\Phi ,$ and this is certainly possible. $\square$

For later use, it will be convenient to choose $C$ to be the unique chamber of $\Sigma$ which is contained in the cone $C_0$ and has the origin $0$ as one vertex. The following lemma presupposes this choice of $C\text{.}$

For each $w\in W,$ let ${\Phi }_w$ be the set of affine roots which are positive on $C$ and negative on $wC\text{.}$

Let $w$ be an element of $W$ such that $C\subset w{C}_0\text{.}$ Then for all $w\prime \in W_0$ we have $${\Phi }_{ww\prime }={\Phi }_w\bigsqcup w{\Phi }_{w\prime }\text{.}$$

		Proof.
	If $\beta \in {\Phi }_w$ is positive on $ww\prime C,$ then $\alpha =w^{-1}\beta$ is negative on $C$ and positive on $w\prime C,$ hence $\alpha \left(0\right)=0$ (because the origin $0$ belongs to the intersection of the closures of $C$ and $w\prime C\text{).}$ Consequently $\alpha$ is negative on $C_0,$ hence $\beta$ is negative on $w{C}_0$ and therefore on $C\text{:}$ contradiction. Hence $\beta$ must be negative on $ww\prime C_0,$ i.e. $\beta \in {\Phi }_{ww\prime }\text{.}$ Next, let $\alpha \in {\Phi }_{w\prime }$ and put $\beta =w\alpha \text{.}$ Then $\beta \left(w{C}_0\right)>0$ and $\beta \left(ww\prime C_0\right)<0,$ hence $\beta$ is positive on $C$ and negative on $ww\prime C,$ i.e. $\beta \in {\Phi }_{ww\prime }\text{.}$ Hence ${\Phi }_w\cup w{\varphi }_{w\prime }\subset {\Phi }_{ww\prime }\text{.}$ Conversely, let $\beta \in {\Phi }_{ww\prime }\text{.}$ If $\beta \left(wC\right)<0,$ then $\beta \in {\Phi }_w\text{;}$ and if $\beta \left(wC\right)>0,$ then $w^{-1}\beta \in {\Phi }_w\text{.}$ This establishes the opposite inclusion. Finally, the sets ${\Phi }_w$ and $w{\Phi }_{w\prime }$ are disjoint, because $\beta \left(wC\right)<0$ for all $\beta \in {\Phi }_w$ and $\beta \left(wC\right)>0$ for all $\beta \in w{\Phi }_{w\prime }\text{.}$ $\square$

Before going any further, it may be helpful to have a concrete example of the set-up so far:

(Root system of type $A_l\text{).}$

Let $A$ be the set of all $x=(x_0,\dots ,x_l)\in {ℝ}^{l+1}$ such that $\sum _{i=0}^l{x}_i=0\text{.}$ The scalar product is $⟨x,y⟩=\sum x_i{y}_i\text{.}$ Let $e_i$ be the $i\text{th}$ coordinate function on $A$ $(e_i\left(x\right)=x_i)$ and let $${\Sigma }_0=\{e_i-e_j:i\ne j\}\text{.}$$ Then ${\Sigma }_0$ satisfies (RS 1)-(RS 5). The Weyl group $W_0$ acts by permuting the coordinates $x_0,\dots ,x_l$ of $x\in A,$ hence is isomorphic to the symmetric group $S_{l+1},$ of order $(l+1)!\text{.}$ We may take the chamber $C_0$ to be $$C_0=\{x\in A:x_0>x_1>\cdots >x_l\}$$ and the simple roots are then $e_{i-1}-e_i$ $(1\le i\le l)\text{.}$ The chamber $C$ is the open simplex $$C=\{x\in A:1+x_l>x_0>x_1>\cdots >x_l\}\text{.}$$

BN-pairs

Let $G$ be a group, $B$ and $N$ subgroups of $G,$ and $R$ a subset of the coset space $N/(B\cap N)\text{.}$ The pair $(B,N)$ is a BN-pair in $G$ (or a Tits system) if it satisfies the following four axioms:

(BN 1)	$G$ is generated by $B$ and $N,$ and $H=B\cap N$ is normal in $N\text{.}$
(BN 2)	$R$ generates the group $W=N/H$ and each $r\in R$ has order $2\text{.}$
(BN 3)	$rBw\subset BwB\cup BrwB,$ for all $r\in R$ and $w\in W\text{.}$
(BN 4)	$rBr\not\subset B,$ for each $r\in R\text{.}$

The group $W=N/H$ is the Weyl group of $(B,N)\text{.}$ The elements of $W$ are cosets of $H,$ and an expression such as $BwB$ is to be read in the usual sense as a product of subsets of $G\text{.}$ If $S$ is a subset of $W,$ then $BSB$ means $\bigcup _{w\in S}BwB\text{.}$

One shows that the set $R$ is uniquely determined by $B$ and $N\text{:}$ namely an element $w\in W$ belongs to $R$ if and only if $B\cup BwB$ is a group. The elements of $R$ are called the distinguished generators of $W\text{.}$

The axioms (BN 1)-(BN 4) have the following consequences. The proofs may be found in [Bou1968].

2.3.1 $G=BWB,$ and the mapping $w\mapsto BwB$ is a bijection of $W$ onto the double coset space $B\backslash G/B\text{.}$

2.3.2 For each subset $S$ of $R$ let $W_S$ be the subgroup of $W$ generated by $S,$ and let $P_S=B{W}_{S}B\text{.}$ Then $P_S$ is a subgroup of $G\text{.}$

2.3.3 The mapping $S\mapsto P_S$ is an inclusion-preserving bijection of the set of subsets of $R$ onto the set of subgroups of $G$ containing $B\text{.}$

2.3.4 If $S,$ $S\prime$ are subsets of $R,$ and $P_S$ is conjugate to $P_{S\prime },$ then $S=S\prime \text{.}$

2.3.5 If $S,$ $S\prime$ are subsets of $R,$ and $w\in W,$ then $$P_{S}w{P}_{S\prime }=B{W}_{S}w{W}_{S\prime }B\text{.}$$

2.3.6 Each $P_S$ is its own normalizer in $G\text{.}$

2.3.7 If $r\in R$ and $w\in W$ and $l\left(rw\right)>l\left(w\right),$ then $BrwB=BrB·BwB\text{.}$ (The length $l\left(w\right)$ of $w\in W$ is defined as the length of a reduced word in the generating set $R,$ just as in (2.1) and (2.2).)

Buildings

Let $G$ be a group with BN-pair $(B,N)$ as in (2.3). We shall assume now that the Weyl group of $(B,N)$ 'is' an affine Weyl group $W$ in the sense of (2.2). More precisely, we shall suppose we are given a reduced irreducible root system ${\Sigma }_0$ in a Euclidean space $A,$ a chamber $C$ for the associated affine root system $\Sigma ,$ and a homomorphism $\nu$ of $N$ onto the affine Weyl group $W$ such that

(1)	$\text{Ker}\left(\nu \right)=H$

so that we may identify (via $\nu \text{)}$ the Weyl group $N/H$ of $(B,N)$ with the Weyl group $W$ of $\Sigma \text{;}$

(2)	under this identification, the distinguished generators of $N/H$ are the reflections in the walls of the chamber $C\text{;}$ in other words, $R=\{w_{\alpha }:\alpha \in \Pi \}\text{.}$

The conjugates of $B$ in $G$ are called Iwahori subgroups of $G\text{.}$ A parahoric subgroup of $G$ is a proper subgroup containing an Iwahori subgroup. From the properties of BN-pairs recalled in (2.3), each parahoric subgroup $P$ is conjugate to $P_S$ for some proper subset $S$ of $R,$ and $S$ is uniquely determined by $P\text{.}$

We set up a one-one correspondence $S⟷F$ between the subsets $S$ of $R$ and the facets $F$ of the chamber $C$ as follows. To a facet $F$ corresponds the set of all $w_{\alpha }\in R$ which fix $F\text{.}$ In this correspondence, $\varnothing ⟷C\text{;}$ $R⟷\varnothing \text{;}$ and $S⟷$ a vertex of $C\iff C$ is a maximal proper subset of $R\text{.}$

If $S⟷F$ we write $P_F$ for $P_S\text{.}$

Each parahoric subgroup $P$ determines uniquely a facet $F\left(P\right)$ of $C\text{:}$ namely $F\left(P\right)=F\iff P$ is conjugate to $P_F\text{.}$

The building associated with the given BN-pair structure on $G$ is the set $ℐ$ of all pairs $(P,x)$ where $P$ is a parahoric subgroup of $G,$ and $x$ is a point in the facet $F\left(P\right)\text{.}$

With each parahoric subgroup $P$ we associate the subset $ℱ\left(P\right)$ of $ℐ,$ where $$ℱ\left(P\right)=\{(P,x):x\in F\left(P\right)\}\text{.}$$ $ℱ\left(P\right)$ is called a facet of $ℐ,$ of type $F\left(P\right),$ In particular, if $P$ is an Iwahori subgroup, $ℱ\left(P\right)$ is a chamber of $ℐ\text{.}$ If $P$ is parahoric, there are only finitely many parahoric subgroups $Q$ containing $P,$ and we define $$\overline{ℱ\left(P\right)}=\bigcup _{Q\supseteq P}ℱ\left(Q\right)\text{.}$$ In this way the building $ℐ$ may be regarded as a geometrical simplicial complex, the $ℱ\left(P\right)$ being the 'open' simplexes and the $\overline{ℱ\left(P\right)}$ the 'closed' simplexes.

The group $G$ acts on $ℐ$ by inner automorphisms: $g(P,x)=(gP{g}^{-1},x)\text{.}$

Apartments

Consider the following subset ${𝒜}_0$ of $ℐ\text{:}$ $${𝒜}_0=\bigcup _{w\in W}\overline{ℱ\left(wB{w}^{-1}\right)}\text{.}$$ Equivalently, ${𝒜}_0$ is the union of the facets $nℱ\left(P\right)$ where $n\in N$ and $P\supset B\text{.}$

Remark. Since we have identified $N/H$ with $W$ by means of the isomorphism induced by $\nu ,$ expressions such as $wP{w}^{-1}$ are meaningful, where $w\in W$ and $P$ is a subgroup of $G$ containing $B\text{.}$ Hence $W$ acts on ${𝒜}_0\text{:}$ $w(P,x)=(wP{w}^{-1},x)=(nP{n}^{-1},x)$ for any $n\in N$ such that $\nu \left(n\right)=w\text{.}$

There exists a unique bisection $j:A\to {𝒜}_0$ such that

(1)	for each facet $F$ of $C$ and each $x\in F,$ $$j\left(x\right)=(P_F,x)\text{;}$$
(2)	$j\circ w=w\circ j$ for all $w\in W\text{.}$

		Proof.
	Let $y\in A\text{.}$ Then $y$ is congruent under $W$ to a unique point $x$ of $\bar{C}\text{:}$ say $y=wx$ where $w\in W$ and $x\in F,$ and $F$ is a facet of the chamber $C\text{.}$ Define $j\left(y\right)$ to be the point $(w{P}_F{w}^{-1},x)\in {𝒜}_0\text{.}$ We have to check that $j$ is well defined. If also $y=w\prime x,$ then $w^{-1}w\prime$ fixes $x$ and therefore, by a property of reflection groups recalled in (2.2), belongs to the subgroup of $W$ generated by the $w_{\alpha },$ $\alpha \in \Pi ,$ which fix $x\text{.}$ In other words, $w^{-1}w\prime \in W_S,$ where $S⟷F\text{.}$ Since $W_S\subset P_S=P_F$ it follows that $w{P}_F{w}^{-1}=w\prime P_F{w\prime }^{-1}\text{.}$ Hence $j$ is well defined, and clearly satisfies (1) and (2). The uniqueness of $j$ is obvious. $\square$

If $g{𝒜}_0={𝒜}_0$ then $j^{-1}\circ \left(g\right|{𝒜}_0)\circ j\in W\text{.}$

		Proof.
	Let ${𝒷}_0$ be the chamber $j\left(C\right)=ℱ\left(B\right)\text{.}$ We have $g{𝒷}_0=n_0{𝒷}_0$ for some $n_0\in N\text{.}$ Hence $g_0=n_0^{-1}g$ fixes ${𝒜}_0$ as a whole and also fixes the chamber ${𝒷}_0$ and all its facets; also it maps each facet to another facet of the same type. Transferring our attention to $A,$ it follows that $\gamma =j^{-1}\circ \left(g_0\right|{𝒜}_0)\circ j$ maps $\underset{\_}{A}$ bijectively onto $A$ and fixes the chamber $C$ and all its facets. Hence $\gamma$ also fixes the adjacent chambers $w_{\alpha }C$ $(\alpha \in \Pi ),$ and so on. It follows that $\gamma$ is the identity mapping and hence that $j^{-1}\circ \left(g\right|{𝒜}_0)\circ j=\nu \left(n_0\right)\in W\text{.}$ $\square$

The subcomplexes $g{𝒜}_0$ of $ℐ,$ as $g$ runs through $G,$ are called the apartments of the building $ℐ\text{.}$ If $𝒜=g{𝒜}_0$ is an apartment, we may transport the Euclidean structure of $A$ to $𝒜$ via the bijection $\left(g\right|{𝒜}_0)\circ j\text{.}$ If also $𝒜=g\prime {𝒜}_0,$ then it follows from (2.4.2) that the two mappings $\left(g\right|{𝒜}_0)\circ j$ and $\left(g\prime \right|{𝒜}_0)\circ j$ differ by an element of $W,$ and hence we have a well-defined structure of Euclidean space on each apartment $𝒜,$ and in particular a distance $d_{𝒜}(x,y)$ defined for $x,y\in 𝒜\text{.}$

Thus the building $ℐ$ may be thought of as obtained by sticking together many copies of the Euclidean space $A\text{.}$

Any two facets of $ℐ$ are contained in a single apartment.

		Proof.
	Consider two facets $ℱ\left(P_1\right)$ and $ℱ\left(P_2\right),$ where $P_1,$ $P_2$ are parahoric subgroups of $G\text{:}$ say $P_i=g_i{P}_{F_i}{g}_i^{-1}$ $(i=1,2),$ where $F_1,$ $F_2$ are facets of the chamber $C$ in $A\text{.}$ By (2.3.1) we can write $g_1^{-1}{g}_2=b_{1}n{b}_2$ with $b_1,b_2\in B$ and $n\in N\text{.}$ If $g=g_1{b}_1,$ then $$P_1=g{P}_{F_1}{g}^{-1},P_2=g\left(n{P}_{F_2}n-1\right)g^{-1}$$ and therefore $ℱ\left(P_1\right)$ and $ℱ\left(P_2\right)$ are both in $g{𝒜}_0\text{.}$ $\square$

$G$ is transitive on the set of pairs $(𝒷,𝒜),$ where $𝒜$ is an apartment and $𝒷$ is a chamber in $𝒜\text{.}$

		Proof.
	We have $𝒷=ℱ\left(g_{0}B{g}_0^{-1}\right)=g_0{𝒷}_0$ for some $g_0\in G,$ where ${𝒷}_0=ℱ\left(B\right)\text{.}$ Hence we may assume that $𝒷={𝒷}_0\text{.}$ If $𝒜=g{𝒜}_0$ contains ${𝒷}_0,$ then $g^{-1}{𝒷}_0$ is a chamber in ${𝒜}_0$ and hence $g^{-1}{𝒷}_0=n{𝒷}_0$ for some $n\in N\text{.}$ So if $g_1=gn$ we have $𝒜=g_1{𝒜}_0$ and ${𝒷}_0=g_1{𝒷}_0\text{.}$ $\square$

Let $𝒜,$ $𝒜\prime$ be, two apartments and let $𝒷$ be a chamber contained in $𝒜\cap 𝒜\prime \text{.}$ Then there exists a unique bijecton $\rho :𝒜\prime \to 𝒜$ such that

(1)	There exists $g\in G$ such that $\rho x=gx$ for all $x\in 𝒜\prime \text{;}$
(2)	$\rho x=x$ for all $x\in 𝒷\text{.}$

Moreover, $\rho x=x$ for all $x\in 𝒜\cap 𝒜\prime ,$ and $d_{𝒜\prime }(x,y)=d_{𝒜}(\rho x,\rho y)$ for all $x,y\in 𝒜\prime \text{.}$

		Proof.
	The existence of $\rho$ follows immediately from (2.4.4). Uniqueness follows from (2.4.2): if ${\rho }_1,{\rho }_2:𝒜\prime \to 𝒜$ both satisfy (1) and (2), then ${\rho }_1\circ {\rho }_2^{-1}$ fixes $𝒜$ as a whole and fixes the chamber $𝒷,$ hence is the identity map. That $d_{𝒜\prime }(x,y)=D_{𝒜}(\rho x,\rho y)$ follows from the definition of the metric on an apartment given after (2.4.2). It remains to show that $\rho x=x$ for all $x\in 𝒜\cap 𝒜\prime \text{.}$ Suppose $ℱ=ℱ\left(P\right)$ is a facet contained in $𝒜\cap 𝒜\prime \text{.}$ By (2.4.4) we may assume that $𝒜\prime ={𝒜}_0,$ $𝒜=g{𝒜}_0$ and $𝒷={𝒷}_0=ℱ\left(B\right)\text{.}$ Since $g{𝒷}_0={𝒷}_0$ it follows that $g$ normalizes $B$ and therefore by (2.3.6) that $g\in B\text{.}$ Since $ℱ\left(P\right)\subseteq {𝒜}_0\cap g{𝒜}_0$ we have $$P=n_1{P}_F{n}_1^{-1}=g\left(n_2{P}_F{n}_2^{-1}\right)g^{-1}$$ for some facet $F$ of $C$ and $n_1,n_2\in N\text{.}$ Since $P_F$ is its own normalizer (2.3.6), we have $n_1^{-1}g{n}_2\in P_F$ and therefore $B{n}_2{P}_F=B{n}_1{P}_F\text{.}$ By (2.3.5), $B{n}_2{W}_{F}B=B{n}_1{W}_{F}B$ where $W_F$ is the subgroup of $W$ which fixes $F\text{.}$ Hence (2.3.1) $n_2{W}_F=n_1{W}_F\text{;}$ since $W_F\subset P_F$ it follows that $n_1{P}_F{n}_1^{-1}=n_2{P}_F{n}_2^{-1},$ hence $ℱ=gℱ=\rho ℱ\text{.}$ $\square$

Retraction of the building onto an apartment

Let $𝒜$ be an apartment and $𝒷$ a chamber in $𝒜\text{.}$ Then there exists a unique mapping $\rho :ℐ\to 𝒜$ such that, for all apartments $𝒜\prime$ containing $𝒷,$ $\rho |𝒜\prime$ is the bisection $𝒜\prime \to 𝒜$ of (2.4.5).

		Proof.
	Let $x\in ℐ\text{.}$ By (2.4.3) there exists an apartment ${𝒜}_1$ containing $x$ and $𝒷\text{.}$ Let ${\rho }_1:{𝒜}_1\to 𝒜$ be the isomorphism of (2.4.5). If also $x$ and $𝒷$ are contained in another apartment ${𝒜}_2,$ then we have also ${\rho }_2:{𝒜}_2\to 𝒜\text{.}$ But then ${\rho }_1^{-1}\circ {\rho }_2:{𝒜}_2\to {𝒜}_1$ is the isomorphism of (2.4.5) for the apartments ${𝒜}_1$ and ${𝒜}_2$ and the chamber $𝒷\text{.}$ Hence ${\rho }_1^{-1}\circ {\rho }_2$ fixes $x,$ that is to say ${\rho }_1\left(x\right)={\rho }_2\left(x\right)\text{.}$ It follows that $\rho$ as in the theorem is well defined, and the uniqueness is obvious. $\square$

The mapping $\rho$ of (2.4.7) is called the retraction of $ℐ$ onto $𝒜$ with centre $𝒷\text{.}$ It has the following properties:

(1)	$\rho x=x$ for all $x\in 𝒜\text{.}$
(2)	For each facet $ℱ$ in $ℐ,$ $\rho |\overline{ℱ}$ is an affine isometry of $\overline{ℱ}$ onto $\overline{\rho ℱ}\text{.}$
(3)	If $x\in \overline{𝒷},$ then ${\rho }^{-1}\left(x\right)=\left\{x\right\}\text{.}$

		Proof.
	(1) and (2) are clear. As to (3), let $ℱ\prime$ be a facet such that $ℱ=\rho ℱ\prime$ is a facet of $𝒷\text{.}$ By (2.4.3) there exists an apartment $𝒜\prime$ containing $𝒷$ and $ℱ\prime ,$ and $\rho |𝒜\prime$ is a bijection of $𝒜\prime$ onto $𝒜$ leaving $𝒷$ fixed. It follows that $ℱ\prime =ℱ\text{.}$ $\square$

(i)	There exists a unique function $d:ℐ\times ℐ\to {ℝ}_{+}$ such that $d|𝒜\times 𝒜$ is the metric $d_{𝒜},$ for each apartment $𝒜$ in $ℐ\text{.}$
(ii)	If $\rho$ is a retraction of $ℐ$ onto an apartment $𝒜$ as in (2.4.6), then $d(\rho \left(x\right),\rho \left(y\right))\le d(x,y)$ for all $x,y\in ℐ\text{.}$
(iii)	$d$ is a $G\text{-invariant}$ metric on $ℐ\text{.}$

		Proof.
	Let $x,y\in {ℐ}_0\text{.}$ By (2.4.3) there is an apartment $𝒜$ containing $x$ and $y\text{.}$ We wish to define $d(x,y)$ to be the Euclidean distance $d_{𝒜}(x,y)$ between $x$ and $y$ in $𝒜\text{.}$ We must therefore show that, if $x,$ $y$ belong also to another apartment $𝒜\prime ,$ then $d_{𝒜}(x,y)=d_{𝒜\prime }(x,y)\text{.}$ Let $𝒷$ be a chamber in $𝒜$ such that $x\in \overline{𝒷}$ and let $𝒷\prime$ be a chamber in $𝒜\prime$ such that $y\in \overline{𝒷}\prime \text{.}$ By (2.4.3) there is an apartment $𝒜″$ containing $𝒷$ and $𝒷\prime \text{.}$ From (2.4.5) we have $$d_{𝒜}(x,y)=d_{𝒜″}(x,y)$$ because $𝒜$ and $𝒜″$ have a chamber in common; and for the same reason $$d_{𝒜\prime }(x,y)=d_{𝒜″}(x,y)\text{.}$$ Hence $d$ is well defined and is $G\text{-invariant}$ (from the definition of $d_{𝒜}\text{).}$ (ii) Let $𝒜\prime$ be an apartment containing $x$ and $y\text{.}$ The image under $\rho$ of the straight line segment joining $x$ to $y$ in $𝒜\prime$ will be a polygonal line in $𝒜$ of the same total length, by (2.4.7)(ii), and endpoints $\rho \left(x\right)$ and $\rho \left(y\right)\text{.}$ Hence the result. (iii) Let $x,y,z\in ℐ$ and let $𝒜$ be an apartment containing $x$ and $y\text{.}$ Let $\rho$ be a retraction of $ℐ$ onto $𝒜\text{.}$ Then $$\begin{array}{ccc}d(x,y)& \le & d(x,\rho \left(z\right))+d(\rho \left(z\right),y)\\ & \le & d(x,z)+d(z,y)\end{array}$$ by (ii) above, since $\rho \left(x\right)=x$ and $\rho \left(y\right)=y\text{.}$ Hence $d$ satisfies the triangle equality, and we have already observed that $d$ is $G\text{-invariant.}$ $\square$

Let $x,y\in ℐ\text{.}$ Then there is a unique geodesic joining $x$ to $y\text{.}$

		Proof.
	Let $𝒜$ be an apartment containing $x$ and $y$ and let $\left[xy\right]$ denote the straight line segment joining $x$ to $y$ in $𝒜\text{.}$ Suppose $z$ lies on a geodesic from $x$ to $y\text{.}$ Then $d(x,z)+d(z,y)=d(x,y)\text{.}$ Let $z_0\in [x,y]$ be such that $d(x,z_0)=d(x,z)$ and hence also $d(z_0,y)=d(z,y)\text{.}$ Let $\rho$ be the retraction of $ℐ$ onto $𝒜$ with centre a chamber $𝒷$ such that $z_0\in \overline{𝒷}\text{.}$ Then we have $$\begin{array}{ccc}d(x,y)& \le & d(x,\rho \left(z\right))+d(\rho \left(z\right),y)\\ & \le & d(x,z)+d(z,y)\text{by (2.4.8)(ii)}\\ & =& d(x,y)\text{.}\end{array}$$ Consequently $$d(x,y)=d(x,\rho \left(z\right))+d(\rho \left(z\right),y)\text{and}d(x,\rho \left(z\right))=d(x,z)\text{.}$$ Since $x,y,\rho \left(z\right)$ are in the same Euclidean space $𝒜$ it follows that $\rho \left(z\right)=z_0\text{.}$ Hence $z\in {\rho }^{-1}\left(z_0\right)$ and therefore $z=z_0$ by (2.4.7)(iii). $\square$

It follows from (2.4.9) that the straight line segment $\left[xy\right]$ lies in all apartments containing $x$ and $y\text{.}$

$ℐ$ is complete with respect to the metric $d\text{.}$

		Proof.
	Let $\left(x_n\right)$ be a Cauchy sequence in $ℐ\text{.}$ If $\rho$ is a retraction of $ℐ$ onto an apartment $𝒜,$ then $\left(\rho \left(x_n\right)\right)$ is a Cauchy sequence in $𝒜,$ by (2.4.8)(ii), and therefore converges to a point $x\in 𝒜\text{.}$ The set of real numbers $d(x,g\left(x\right)),$ where $g\in G$ is such that $x\ne gx,$ has a positive lower bound $\mu ,$ and we shall have $$d(\rho \left(x_n\right),x)<\frac{1}{4}\mu ,d(x_n,x_{n+1})<\frac{1}{4}\mu$$ for all sufficiently large $n\text{.}$ Now each $\rho \left(x_n\right)$ is of the form $g_n{x}_n$ for some $g_n\in G\text{.}$ Put $y_n=g_n^{-1}x\text{.}$ Then $$\begin{array}{ccc}d(y_n,y_{n+1})& \le & d(y_n,x_n)+d(x_n,x_{n+1})+d(x_{n+1},y_{n+1})\\ & =& d(x,\rho \left(x_n\right))+d(x_n,x_{n+1})+d(\rho \left(x_{n+1}\right),x)\\ & <& \frac{3}{4}\mu <\mu \end{array}$$ for all large $n\text{.}$ Hence $y_n=y_{n+1}=\dots =y$ and we have $$d(x_n,y)=d(x_n,y_n)=d(\rho \left(x_n\right),x)\to 0$$ as $n\to \infty \text{.}$ Hence $x_n\to y\text{.}$ $\square$

Fixed point theorem

A subset $X$ of $ℐ$ is convex if $x,y\in X\Rightarrow \left[xy\right]\subset X\text{.}$

Let $X$ be a bounded non-empty subset of $ℐ\text{.}$ Then the group of isometries $\gamma$ of $ℐ$ such that $\gamma \left(X\right)\subseteq X$ has a fixed point in the closure of the convex hull of $X\text{.}$

For the proof we shall require the following lemma:

Let $x,y,z\in ℐ$ and let $m$ be the midpoint of $\left[xy\right]\text{.}$ Then $$d{(z,x)}^2+d{(z,y)}^2\ge 2d{(z,m)}^2+\frac{1}{2}d{(x,y)}^2\text{.}$$

		Proof.
	If $x,y,z$ are all in the same apartment $𝒜,$ this is true with equality, by elementary geometry (take $z$ as origin). In the general case, let $𝒜$ be an apartment containing $x$ and $y$ and choose a chamber $𝒷$ in $𝒜$ such that $m\in \overline{𝒷}\text{.}$ Retract $ℐ$ onto $𝒜$ with centre $𝒷\text{.}$ Then $$\begin{array}{ccc}d{(z,x)}^2+d{(z,y)}^2& \ge & d{(\rho \left(z\right),x)}^2+d{(\rho \left(z\right),y)}^2\\ & =& 2d{(\rho \left(z\right),m)}^2+\frac{1}{2}d{(x,y)}^2\\ & =& 2d{(z,m)}^2+\frac{1}{2}d{(x,y)}^2\text{.}\end{array}$$ $\square$

		Proof of 2.4.11.
	Let $\delta \left(X\right)=\underset{x,y\in X}{\text{sup}}d(x,y)$ be the diameter of $X\text{.}$ Choose a real number $k\in (0,1)$ and let $fX$ be the set of all $m\in ℐ$ such that $m$ is the midpoint of $\left[xy\right]$ with $x,y\in X$ and $d(x,y)\ge k·\delta \left(X\right)\text{.}$ Clearly $fX$ is non-empty and is contained in the convex hull of $X\text{.}$ If $m\in fX$ and $z\in X,$ then by (2.4.12) $$\begin{array}{ccc}d{(z,m)}^2& =& \frac{1}{2}d{(z,x)}^2+\frac{1}{2}d{(z,y)}^2-\frac{1}{4}d{(x,y)}^2\\ & \le & \frac{1}{2}\delta {\left(X\right)}^2+\frac{1}{2}\delta {\left(X\right)}^2-\frac{1}{4}{k}^2\delta {\left(X\right)}^2\\ & =& (1-\frac{1}{4}{k}^2)\delta {\left(X\right)}^2\end{array}$$ so that $$\begin{array}{cc}d(z,m)\le k_1\delta \left(X\right)& \text{(1)}\end{array}$$ for some $k_1<1\text{.}$ If now $m\in fX$ and $z\in fX,$ then again $m$ is the midpoint of a segment $\left[xy\right]$ with $x,y\in X$ and $d(x,y)\ge k\delta \left(X\right),$ so that $$\begin{array}{ccc}d{(z,m)}^2& \le & \frac{1}{2}d{(z,x)}^2+\frac{1}{2}d{(z,y)}^2-\frac{1}{4}d{(x,y)}^2\\ & \le & (1-\frac{1}{4}{k}^2)\delta {\left(X\right)}^2-\frac{1}{4}{k}^2\delta {\left(X\right)}^2\\ & =& (1-\frac{1}{2}{k}^2)\delta {\left(X\right)}^2\end{array}$$ so that $$\begin{array}{cc}\delta \left(fX\right)\le k_2\delta \left(X\right)& \text{(2)}\end{array}$$ for some $k_2<1\text{.}$ From (2) it follows that $$\begin{array}{cc}\delta \left(f^{n}X\right)\to 0\text{as}\hspace{0.17em}n\to \infty \text{.}& \text{(3)}\end{array}$$ For each $n,$ pick $x_n\in f^{n}X\text{.}$ Then by (1) and (2) $$d(x_n,x_{n+1})\le k_1\delta \left(f^{n}X\right)\le k_1{k}_2^n\delta \left(X\right)\text{.}$$ Hence $\left(x_n\right)$ is a Cauchy sequence in $ℐ,$ hence by (2.4.10) converges to some $x\in ℐ\text{.}$ Clearly $x$ lies in the closure of the convex hull of $X\text{.}$ Now let $\gamma$ be an isometry of $ℐ$ such that $\gamma X\subseteq X\text{.}$ Then $\gamma f^{n}X\subseteq f^{n}X$ for all $n\text{.}$ Let $x_n^{\prime }=\gamma \left(x_n\right)\text{.}$ Then $\left(x_n^{\prime }\right)$ is a Cauchy sequence with $x_n^{\prime }\in f^{n}X$ for all $n$ and converges to $\gamma x\text{.}$ But by (3) $d(x_n,x_n^{\prime })\to 0$ as $n\to \infty \text{.}$ Hence $\gamma x=x\text{.}$ Hence $x$ is a fixed point of the group of isometries $\gamma$ of $ℐ$ such that $\gamma X\subseteq X\text{.}$ $\square$

A subset $M$ of $G$ is said to be bounded if $MX$ is bounded for all bounded subsets $X$ of $ℐ\text{.}$

$M$ is bounded $\iff$ $M$ intersects only finitely many double cosets $BwB$ $(w\in W)\text{.}$

		Proof.
	Let $\rho$ be the retraction of $ℐ$ onto the apartment ${𝒜}_0$ with centre ${𝒷}_0=ℱ\left(B\right)\text{.}$ Then $X\subset ℐ$ is bounded $\iff$ $\rho X$ is bounded $\iff$ $\rho X$ is contained in a finite union of closed chambers $\overline{𝒷}$ of ${𝒜}_0\text{.}$ Hence $M$ is bounded $\iff$ $M𝒷$ is bounded, for each chamber $𝒷\subset {𝒜}_0\iff M{𝒷}_0$ is bounded. If $m\in M,$ define $w_m\in W$ by the relation $m\in B{w}_{m}B\text{.}$ Then $M{𝒷}_0$ is bounded $\iff \bigcup _{m\in M}{w}_m{𝒷}_0$ is bounded $\iff$ the set $\{w_m:m\in M\}$ is finite $\iff M$ intersects only finitely many $BwB\text{.}$ $\square$

A subgroup $\Gamma$ of $\Gamma$ is bounded $\iff$ $\Gamma$ is contained in a parahoric subgroup.

		Proof.
	$\Rightarrow$ Let $x\in ℐ\text{.}$ Then $X=\Gamma x$ is bounded. Applying the fixed point theorem (2.4.11), we see that there exists $y\in ℐ$ such that $\Gamma y=y\text{.}$ If $y$ lies in the facet $ℱ\left(P\right),$ then $\Gamma$ normalizes the parahoric subgroup $P$ and hence $\Gamma \subseteq P$ by (2.3.6). $\Leftarrow$ Replacing $\Gamma$ by a conjugate, we may assume that $\Gamma \subset P_S,$ where $S$ is a proper subset of $R=\{w_{\alpha }:\alpha \in \Pi \}\text{.}$ Now $P_S=B{W}_{S}B,$ and $W_S$ is finite because $S\ne R\text{.}$ Hence $\Gamma$ is bounded by (2.4.13). $\square$

Every bounded subgroup of $G$ is contained in a maximal bounded subgroup. The maximal bounded subgroups are precisely the maximal parahoric subgroups of $G,$ and form $l+1$ conjugacy classes, corresponding to the vertices of the chamber $C\text{.}$

Groups with affine root structure

(a) Universal Chevalley groups

We retain the notation of (2.1) and (2.2). Let $$L=\{u\in A^{*}:u\left(a^{\vee }\right)\in \mathbb{Z}\hspace{0.17em}\text{for all}\hspace{0.17em}a\in {\Sigma }_0\}\text{.}$$ Then $L$ is a lattice of rank $l$ $(=\text{dim}\hspace{0.17em}A)$ in $A^{*},$ containing ${\Sigma }_0\text{.}$

Let $𝓀$ be any field, and ${𝓀}^{+},{𝓀}^{\times }$ the additive and multiplicative groups of $𝓀\text{.}$ Associated with the pair $({\Sigma }_0,𝓀)$ there is a universal Chevalley group $G=G({\Sigma }_0,𝓀)\text{.}$ We shall not go into the details of the construction of $G,$ for which we refer to Steinberg [Ste1967]. All we shall say here is that $G$ is generated by elements $u_a\left(\xi \right)$ $(a\in {\Sigma }_0,\xi \in 𝓀)$ and contains elements $h\left(\chi \right)$ $(\chi \in \text{Hom}(L,{𝓀}^{\times }))$ with various relations between the $u\text{'s}$ and the $h\text{'s.}$ In particular, each $u_a$ is an injective homomorphism of ${𝓀}^{+}$ into $G,$ and $h$ is an injective homomorphism of $\text{Hom}(L,{𝓀}^{\times })$ into $G\text{.}$

If ${\Sigma }_0$ is the root system of type $A_l$ (2.2.3) then the universal Chevalley group $G({\Sigma }_0,𝓀)$ is the special linear group ${SL}_{l+1}\left(𝓀\right)\text{.}$ In this case the $u_a\left(\xi \right)$ and $h\left(\chi \right)$ may be taken as follows:

If $a=e_i-e_j$ $(i\ne j)$ then $u_a\left(\xi \right)=1+\xi E_{ij},$ where $1$ is the unit matrix and $E_{ij}$ is the matrix with $1$ in the $(i,j)$ place and $0$ elsewhere. $h\left(\chi \right)$ is the diagonal matrix with $\chi \left(e_0\right),\dots ,\chi \left(e_l\right)$ in order down the diagonal.

Now suppose that the field $𝓀$ carries a discrete valuation $v,$ i.e. a surjective homomorphism $v:{𝓀}^{\times }\to \mathbb{Z}$ such that $v(\xi +\eta )\ge \text{min}(v\left(\xi \right),v\left(\eta \right))$ for all $\xi ,\eta \in 𝓀$ (conventionally $v\left(0\right)=+\infty \text{).}$ We consider certain subgroups of the Chevalley group $G=G({\Sigma }_0,𝓀)\text{.}$

First, let $Z$ be the subgroup consisting of the $h\left(\chi \right)$ and let $N$ be the normalizer of $Z\text{.}$ Then there is a canonical isomorphism $N/Z\to W_0,$ and one can show that this lifts uniquely to a homomorphism $\nu$ of $N$ onto the affine Weyl group $W\text{.}$ The restriction of $\nu$ to $\mathbb{Z}$ is described as follows. If $\chi \in \text{Hom}(L,{𝓀}^{\times })$ then $v\circ \chi$ is a homomorphism $L\to \mathbb{Z},$ which defines a linear form $A^{*}\to ℝ,$ hence a vector in $A\text{.}$ Then $\nu \left(h\left(\chi \right)\right)$ is translation by this vector, and this translation belongs to $W\text{.}$ We have $\nu \left(Z\right)=T$ and $Z={\nu }^{-1}\left(T\right),$ where $T$ is as before the translation subgroup of $W\text{.}$ The kernel $H$ of $\nu$ is the set of all $h\left(\chi \right)$ such that $\chi \left(L\right)$ is contained in the group of units of $𝓀\text{.}$

Next, for each $\alpha \in \Sigma$ we have a subgroup $U_{\alpha }$ of $G,$ defined as follows. If $\alpha =a+k,$ then $$U_{a+k}=\{u_a\left(\chi \right):\xi \in 𝓀\hspace{0.17em}\text{and}\hspace{0.17em}v\left(\xi \right)\ge k\}\text{.}$$ If $\Phi$ is any subset of $\Sigma ,$ we denote by $U_{\Phi }$ the subgroup of $G$ generated by the $U_{\alpha }$ such that $\alpha \in \Phi ,$ and by $P_{\Phi }$ the subgroup generated by $U_{\Phi }$ and $H\text{.}$ Then the triple $(N,\nu ,{\left(U_{\alpha }\right)}_{\alpha \in \Sigma })$ has the following properties, for all $n\in N$ and $\alpha ,\beta \in \Sigma \text{:}$

(I)	$n{U}_{\alpha }{n}^{-1}=U_{\nu \left(n\right)\alpha }\text{.}$

From this it follows that $H$ normalizes each $U_{\alpha },$ hence each $U_{\Phi }\text{.}$ Consequently $$P_{\Phi }=H·U_{\Phi }=U_{\Phi }·H\text{.}$$

(II)	If $k$ is a positive integer, then $U_{\alpha +k}$ is a proper subgroup of $U_{\alpha },$ and $\bigcap _{k\ge 0}{U}_{\alpha +k}=\left\{1\right\}\text{.}$

From this it follows that for each $a\in {\Sigma }_0,$ the union of the $U_{a+r},$ $r\in \mathbb{Z}$ is a group, which we denote by $U_{\left(a\right)}\text{.}$ (In the present situation $U_{\left(a\right)}=u_a\left({𝓀}^{+}\right)\text{.)}$

The next three properties relate to a pair of subgroups $U_{\alpha }$ and $U_{\beta }$ in the following situations: (III) $\beta +\alpha$ is a positive constant; (IV) $\beta +\alpha =0\text{;}$ (V) neither $\beta +\alpha$ nor $\beta -\alpha$ is constant (i.e. the hyperplanes $h_{\alpha },$ $h_{\beta }$ are not parallel).

(III)	If $\beta =-\alpha +k$ with $k>0,$ then $P_{\{\alpha ,\beta \}}=U_{\alpha }H{U}_{\beta }\text{.}$
(IV)	$P_{\{\alpha ,-\alpha \}}(U_{\alpha }·{\nu }^{-1}\left(w_{\alpha }\right)·U_{\alpha })\cup \left(U_{\alpha }H{U}_{-\alpha +1}\right)\text{.}$

For $\alpha ,\beta \in \Sigma$ let $[\alpha ,\beta ]$ denote the set of all $\gamma \in \Sigma$ of the form $m\alpha +n\beta$ with $m,$ $n$ positive integers, and let $[U_{\alpha },U_{\beta }]$ be the subgroup of $G$ generated by all commutators $(u,v)=uv{u}^{-1}{v}^{-1}$ with $u\in U_{\alpha }$ and $v\in U_{\beta }\text{.}$ Then from Chevalley's commutator relations we have

(V)	If $h_{\alpha }$ and $h_{\beta }$ are not parallel, then $[U_{\alpha },U_{\beta }]\subseteq U_{[\alpha ,\beta ]}\text{.}$

Let $U^{+}=⟨U_{\left(a\right)}:a\in {\Sigma }_0^{+}⟩$ be the subgroup generated by the $U_{\left(a\right)}$ with $a$ positive, and likewise $U^{-}=⟨U_{\left(a\right)}:a\in {\Sigma }_0^{-}⟩\text{.}$ Then

(VI)	$U^{+}\cap Z{U}^{-}=\left\{1\right\}\text{.}$

Finally,

(VII)

$G$ is generated by $N$ and the $U_{\alpha }$ $(\alpha \in \Sigma )\text{.}$

(b) Simply-connected simple algebraic groups

As before let $𝓀$ be a field with a discrete valuation $v,$ and assume now that $𝓀$ is complete with respect to $v,$ with perfect residue field.

Let $𝒢$ be a simply-connected simple linear algebraic group defined over $𝓀\text{.}$ Let $𝒮$ be a maximal $𝓀\text{-split}$ torus in $𝒢,$ and let $𝒵,𝒩$ be the centralizer and normalizer respectively of $𝒮$ in $𝒢\text{.}$ Let $G=𝒢\left(𝓀\right)$ be the group of $𝓀\text{-rational}$ points of $𝒢\text{.}$ Then one of the main results of the structure theory of Bruhat and Tits ([BTi1966], [BTi1967]) is:

There exists a reduced irreducible root system ${\Sigma }_0,$ subgroups $N$ and $U_{\alpha }$ $(\alpha \in \Sigma )$ of $G,$ and a surjective homomorphism $\nu :N\to W$ such that the triple $(N,\nu ,{\left(U_{\alpha }\right)}_{\alpha \in \Sigma })$ satisfies (I)-(VII) above.

In fact one takes $N=𝒩\left(𝓀\right),$ $Z=𝒵\left(𝓀\right),$ and $H=\text{Ker}\left(\nu \right)$ is the set of all $z\in 𝒵\left(𝓀\right)$ such that $v\left(\chi \left(z\right)\right)=0$ for all characters $\chi$ of $𝒵\text{.}$ $W_0$ is the relative Weyl group of $𝒢$ with respect to $𝒮,$ but ${\Sigma }_0$ is not in general the relative root system, even if the latter is reduced. But every $a\in {\Sigma }_0$ is proportional to a root of $G$ relative to $𝒮,$ and conversely.

If $𝒢$ is split over $𝓀$ (i.e. if $𝒮$ is a maximal torus of $𝒢\text{)}$ then we are back in the situation of (a) above.

If $𝓀$ is a $p\text{-adic}$ field (i.e. if the residue field of $𝓀$ is finite) then $G=𝒢\left(𝓀\right)$ inherits a topology from $𝓀$ for which $G$ is a locally compact topological group. If $K$ is a suitably chosen maximal compact subgroup of $G,$ then as we shall see later (3.3.7) the algebra $ℒ(G,K)$ is commutative, and the theory of zonal spherical functions on $G$ relative to $K$ is almost entirely a consequence of the properties (I)-(VII) listed above. We shall therefore take these properties as a set of axioms:

Let ${\Sigma }_0$ be a reduced irreducible root system and let $\Sigma ,$ $W$ be the associated affine root system and affine Weyl group, as in (2.2). An affine root structure of type ${\Sigma }_0$ on a group $G$ is a triple $(N,\nu ,{\left(U_{\alpha }\right)}_{\alpha \in \Sigma }),$ where $N$ and the $U_{\alpha }$ are subgroups of $G,$ and $\nu$ is a homomorphism of $N$ onto $W,$ satisfying (I)-(VII) above.

Cartan and Iwasawa decompositions

In this section $G$ is a group endowed with an affine root structure $(N,\nu ,{\left(U_{\alpha }\right)}_{\alpha \in \Sigma })$ of type ${\Sigma }_0$ (2.5.3). We shall develop some consequences of the axioms (I)-(VII).

Let $\Phi$ be a subset of $\Sigma$ such that

(1)	if $\alpha ,\beta \in \Phi$ and $\alpha +\beta \in \Sigma ,$ then $\alpha +\beta -k\in \Phi$ for some $k\ge 0\text{;}$
(2)	if $\alpha ,\beta \in \Phi$ and $\alpha \ne \beta ,$ then the hyperplanes $h_{\alpha },$ h_{\beta }$$ are not parallel.

Let $U_{\Phi }$ be the subgroup of $G$ generated by the $U_{\alpha },$ $\alpha \in \Phi \text{.}$ Then every element of $U_{\Phi }$ can be written uniquely in the form $u_1\cdots u_r,$ where $u_i\in U_{{\alpha }_i}$ $(1\le i\le r)$ and ${\alpha }_1,\dots ,{\alpha }_r$ are the elements of $\Phi$ in any fixed order.

		Proof.
	Let $a_l$ be the gradient of ${\alpha }_i\text{.}$ Then $a_1,\dots ,a_r$ are all distinct by (2) and ${\Phi }_0=\{a_1,\dots ,a_r\}$ is a closed subset of ${\Sigma }_0$ such that ${\Phi }_0\cap -{\Phi }_0=\varnothing$ (by (1) and (2)). Hence (2.1.2) there exists $w\in W_0$ such that $w{\Phi }_0\subseteq {\Sigma }_0^{+}\text{.}$ Choosing $n\in N$ such that $\nu \left(n\right)=w,$ we have $n{U}_{\Phi }{n}^{-1}=U_{w\Phi }$ by (I); hence we may assume without loss of generality that ${\Phi }_0\subseteq {\Sigma }_0^{+}\text{.}$ Assume first that $(a_1,\dots ,a_r)$ in this order is a subsequence of the sequence $(b_1,\dots ,b_n)$ of (2.1.1). Let ${\Phi }_1=\{{\alpha }_2,\dots ,{\alpha }_r\}\text{.}$ If $m,n$ are positive integers and $i>1,$ then $m{a}_1+n{a}_i\ne a_1,$ hence $U_{[{\alpha }_1,{\alpha }_i]}\subseteq U_{{\Phi }_1}\text{.}$ Now use the commutator axiom (V): if $u_1\in U_{{\alpha }_1}$ and $u_i\in U_{{\alpha }_i}$ $(i>1)$ we have $u_1{u}_i{u}_1^{-1}{u}_i^{-1}\in U_{[{\alpha }_1,{\alpha }_i]}$ and therefore $u_1{u}_i{u}_1^{-1}\in U_{{\Phi }_1}\text{.}$ This shows that $U_{{\Phi }_1}$ is normal in $U_{\Phi },$ and hence that $U_{\Phi }=U_{{\alpha }_1}{U}_{{\Phi }_1}\text{.}$ By induction on $r=\text{Card}\left(\Phi \right)$ we conclude that $$U_{\Phi }=U_{{\alpha }_1}{U}_{{\alpha }_2}\cdots U_{{\alpha }_r}\text{.}$$ Next we shall show that every $u\in U_{\Phi }$ is uniquely of the form $u=u_1\cdots u_r$ with $u_i\in U_{{\alpha }_i}$ $(1\le i\le r)\text{.}$ Again by induction, it is enough to show that $u_1$ is uniquely determined by $u,$ and for this it is enough to show that $U_{{\alpha }_1}\cap U_{{\Phi }_1}=\left\{1\right\}\text{.}$ By (2.1.1) we can assume that $a_1$ is negative and $a_2,\dots ,a_r$ ar are positive. Then $U_{{\alpha }_1}\subset U^{-}$ and $U_{{\Phi }_1}\subset U^{+},$ hence the result follows from (VI). This proves (2.6.1) for the particular ordering chosen. For an arbitrary ordering of $\Phi$ it follows from (2.6.2) below, whose proof we leave to the reader. $\square$

Let $X$ be a group with subgroups $X_1,\dots ,X_r$ such that

(1)	$X=X_1{X}_2\cdots X_r$ with uniqueness of expression;
(2)	$X_i{X}_{i+1}\cdots X_r$ is a normal subgroup of $X,$ for $i=1,2,\dots ,r\text{.}$

If $p$ is any permutation of $\{1,2,\dots ,r\},$ then $X=X_{p\left(1\right)}{X}_{p\left(2\right)}\cdots X_{p\left(r\right)}$ with uniqueness of expression.

$U^{+}=\prod _{a\in {\Sigma }_0^{+}}{U}_{\left(a\right)}$ with uniqueness of expression, the positive roots being taken in any fixed order. Likewise $U^{-}=\prod _{a\in {\Sigma }_0^{-}}{U}_{\left(a\right)}\text{.}$

		Proof.
	Arrange the positive roots in a sequence $(b_1,\dots ,b_n)$ satisfying (2.1.1). Follow the proof of (2.6.1) to show that $U^{+}=\prod _{i=1}^n{U}_{b_i}$ with uniqueness of expression, and then use (2.6.2). $\square$

If $S$ is any subset of $A,$ let $\left(S\right)$ denote the set of all $\alpha \in \Sigma$ which are $\ge 0$ on $S\text{.}$ Then the subgroups $U_{\left(S\right)},$ $P_{\left(S\right)}$ are defined.

Let $a\in {\Sigma }_0\text{.}$ If there is an affine root with gradient $a$ belonging to $\left(S\right),$ there is a least such affine root, say $\alpha :$ $\alpha \in \left(S\right)$ and $\alpha -1\notin \left(S\right)\text{.}$ Let $U_{\left(S\right),a}$ a denote the corresponding subgroup $U_{\alpha }$ of $G\text{.}$ If there is no affine root with gradient $a$ belonging to $\left(S\right),$ define $U_{\left(S\right),a}$ to be $\left\{1\right\}\text{.}$ Let $U_{\left(S\right)}^{+}$ (resp. $U_{\left(S\right)}^{-}\text{)}$ be the group generated by the $U_{\left(S\right),a}$ with $a\in {\Sigma }_0^{+}$ (resp. $a\in {\Sigma }_0^{-}\text{).}$

Suppose $S$ has non-empty interior. Then $$P_{\left(S\right)}=U_{\left(S\right)}^{+}H{U}_{\left(S\right)}^{-}\text{.}$$

		Proof.
	By (2.6.1), $U_{\left(S\right)}^{\pm }=\prod _{a\in {\Sigma }_0^{\pm }}{U}_{\left(S\right),a},$ the product being taken in any order. Since $P_{\left(S\right)}$ is generated by the $U_{\left(S\right),a}$ and $H$ it is enough to show that $U_{\left(S\right)}^{+}H{U}_{\left(S\right)}^{-}$ is a group. For this it is enough to show that $U_{\left(S\right)}^{+}H{U}_{\left(S\right)}^{-}=U_{\left(S\right)}^{-}H{U}_{\left(S\right)}^{+}\text{.}$ Arrange the positive roots in a sequence $(b_1,\dots ,b_n)$ satisfying (2.1.1) Let $U_{\pm r}=U_{\left(S\right),\pm b_r}$ $(1\le r\le n),$ and let $V_r^{+}$ (resp. $V_r^{-}\text{)}$ be the subgroup of $G$ generated by $U_{-1},\dots ,U_{-r},U_{r+1},\dots ,U_n$ (resp. by $U_1,\dots ,U_r,U_{-r-1},\dots ,U_{-n}\text{).}$ Since $S$ has non-empty interior it follows from (III) that $U_{r}H{U}_{-r}$ is a group and therefore $$U_{r}H{U}_{-r}=U_{-r}H{U}_r\text{.}$$ Hence, by (2.6.1), we have $$V_{r-1}^{+}H{V}_{r-1}^{-}=V_r^{+}H{V}_r^{-}$$ for $r=1,2,\dots ,n\text{.}$ Consequently $$U_{\left(S\right)}^{+}H{U}_{\left(S\right)}^{-}=V_0^{+}H{V}_0^{-}=V_n^{+}H{V}_n^{-}=U_{\left(S\right)}^{-}H{U}_{\left(S\right)}^{+}\text{.}$$ $\square$

If $\alpha \in \Sigma$ then $U_{\alpha }\subseteq P_{\left(S\right)}\Rightarrow \alpha \in \left(S\right)\text{.}$

		Proof.
	Suppose $U_{\alpha }\subseteq P_{\left(S\right)}\text{.}$ Let $u\in U_{\alpha },$ then $u=u^{+}h{u}^{-}$ with $u^{\pm }\in U_{\left(S\right)}^{\pm }$ and $h\in H\text{.}$ Suppose, to fix the ideas, that the gradient $a$ of $\alpha$ is a positive root. Then ${\left(u^{+}\right)}^{-1}u=h{u}^{-}\in U^{+}\cap Z{U}^{-}=\left\{1\right\}$ by (VI). Hence $u=u^{+},$ i.e. $u\in U_{\left(S\right)}^{+}\text{.}$ By (2.6.1) and (2.6.3) it follows that $u\in U_{\left(S\right),a}\text{.}$ Hence $U_{\alpha }\subseteq U_{\left(S\right),\alpha }$ and therefore by (II) we have $\alpha \in \left(S\right)\text{.}$ The reverse implication is obvious. $\square$

We shall now show that the group $G$ carries two BN-pair structures, one with Weyl group $W$ and the other with Weyl group $W_0$ (2.6.8) and (2,6.9). To establish these results we need a couple of preliminary lemmas.

(1)	If $\alpha \in \Sigma ,$ then $U_{-\alpha }-U_{-\alpha +1}\subset U_{\alpha }·{\nu }^{-1}\left(w_{\alpha }\right)·U_{\alpha }\text{.}$
(2)	If $a\in {\Sigma }_0,$ then $U_{-a}\subset U_a·{\nu }_0^{-1}\left(w_a\right)·U_a\text{.}$

(Here ${\nu }_0$ is the epimorphism $N\to W_0$ obtained by composing $\nu :N\to W$ with the epimorphism $W\to W_0$ defined by the semi-direct decomposition $W=T·W_0\text{.)}$

		Proof.
	(1) Let $u\in U_{-\alpha },$ then by (IV) we have $$u\in U_{\alpha }·{\nu }^{-1}\left(w_{\alpha }\right)·U_{\alpha }\cup U_{\alpha }H{U}_{-\alpha +1}\text{.}$$ Suppose that $u\in U_{\alpha }H{U}_{-\alpha +1}\text{.}$ Then $u=u_{1}h{u}_2$ with $u_1\in U_{\alpha },$ $u_2\in U_{-\alpha +1},$ $h\in H\text{.}$ Hence $u{u}_2^{-1}=u_{1}h\in U_{\alpha }H\cap U_{-\alpha }=\left\{1\right\}$ by (VI). Consequently $u=u_2\in U_{-\alpha +1},$ whence the result. (2) Let $u\in U_{-a},$ $u\ne 1\text{.}$ Then by (II) there exists an affine root $\alpha$ with gradient $a$ such that $u\in U_{-\alpha }-U_{-\alpha +1}\text{.}$ Hence by (1) just proved we have $$u\in U_{\alpha }·{\nu }^{-1}\left(w_{\alpha }\right)·U_{\alpha }\subset U_a·{\nu }_0^{-1}\left(w_a\right)·U_a\text{.}$$ $\square$

Let $B=P_{\left(C\right)}$ and let $B_0^{\pm }=Z{U}^{\pm }\text{.}$ Since $Z$ normalizes $U^{+},$ $B_0^{+}$ is a subgroup of $G,$ and is the semidirect product of $Z$ and $U^{+}$ by axiom (VI). Similarly for $B_0^{-}\text{.}$

(1)	If $\alpha \in \Pi$ and $n_{\alpha }\in {\nu }^{-1}\left(w_{\alpha }\right),$ then $B{n}_{\alpha }B=B{n}_{\alpha }{U}_{\alpha }\text{.}$
(2)	If $a\in {\Pi }_0$ and $n_a\in {\nu }_0^{-1}\left(w_a\right),$ then $B_0^{+}{n}_a{B}_0^{+}=B_0^{+}{n}_a{U}_a\text{.}$

		Proof.
	(1) Since $\alpha \left(C\right)>0$ we have $U_{\alpha }\subset B,$ hence $B{n}_{\alpha }{U}_{\alpha }\subseteq B{n}_{\alpha }B\text{.}$ Conversely, if $\beta \in \left(C\right)$ and $\beta \ne \alpha ,$ then since $h_{\alpha }$ is the only hyperplane separating $C$ and $w_{\alpha }C,$ it follows that $\beta$ is positive on $w_{\alpha }C,$ hence that $U_{\beta }\subset P_{\left(w_{\alpha }C\right)}=n_{\alpha }^{-1}B{n}_{\alpha }\text{.}$ Consequently $B{n}_{\alpha }{U}_{\beta }=B{n}_{\alpha }$ for all such $\beta ,$ and therefore $B{n}_{\alpha }B=B{n}_{\alpha }{U}_{\alpha }\text{.}$ (2) Let $U^a$ be the group generated by the $U_b$ with $b\in {\Sigma }_0^{+}$ and $b\ne a\text{.}$ By (2.6.3) we have $U^{+}=U^a{U}_a\text{.}$ Hence, since $w_a$ permutes ${\Sigma }_0^{+}-\left\{a\right\},$ $$B_0^{+}{n}_a{B}_0^{+}=B_0^{+}{n}_{a}Z{U}^a{U}_a=B_0^{+}Z{U}^a{n}_a{U}_a=B_0^{+}{n}_a{U}_a\text{.}$$ $\square$

Let $B=P_{\left(C\right)}\text{.}$ Then $(B,N)$ is a BN-pair in $G\text{.}$ We have $B\cap N=H,$ so that $\nu$ induces an isomorphism of $N/(B\cap N)$ onto the affine Weyl group $W\text{.}$ The distinguished generators of $N/(B\cap N)$ correspond under this isomorphism to the reflections in the walls of the chamber $C\text{.}$

		Proof.
	(1) Let $n\in B\cap N\text{.}$ Then $n{P}_{\left(C\right)}{n}^{-1}=P_{\left(C\right)},$ so that $P_{\left(wC\right)}=P_{\left(C\right)}$ where $w=\nu \left(n\right)\text{.}$ Suppose that $w\ne 1,$ so that $wC\ne C\text{.}$ Then there exists a hyperplane $h_{\alpha }$ $(\alpha \in \Sigma )$ separating $C$ and $wC,$ hence an affine root $\alpha$ which is positive on $wC$ and negative on $C\text{.}$ Then $U_{\alpha }\subset P_{\left(wC\right)}=P_{\left(C\right)},$ and since $\alpha \notin \left(C\right)$ this contradicts (2.6.5). Hence $w=1$ and therefore $n\in H\text{.}$ Hence $B\cap N=H\text{.}$ (2) We have to verify the axioms (BN 1)-(BN 4) of (2.3). Of these, (BN 1) follows immediately from (I) and (VII). (BN 2) is clear: we define $R$ to be the set $\{{\nu }^{-1}\left(w_{\alpha }\right):\alpha \in \Pi \}\text{.}$ So is (BN 4), because we have already shown that $P_{\left(wC\right)}\ne P_{\left(C\right)}$ if $w\ne 1\text{.}$ As to (BN 3), let $w\in W$ and $\alpha \in \Pi ,$ and let $n\in {\nu }^{-1}\left(w\right),$ $n_{\alpha }\in {\nu }^{-1}\left(w_{\alpha }\right)\text{.}$ Then $$n_{\alpha }Bn\subset B{n}_{\alpha }Bn=B{n}_{\alpha }{U}_{\alpha }n$$ by (2.6.7). Suppose first that $\alpha \left(wC\right)>0\text{.}$ Then $U_{w^{-1}\alpha }\subset B$ and so $$B{n}_{\alpha }{U}_{\alpha }n=B{n}_{\alpha }n{U}_{w^{-1}\alpha }\subset B{n}_{\alpha }nB\text{.}$$ If on the other hand $\alpha \left(wC\right)<0,$ then $$B{n}_{\alpha }{U}_{\alpha }n=B{U}_{-\alpha }{n}_{\alpha }n$$ and by axiom (IV) we have $$U_{-\alpha }\subset U_{\alpha }{n}_{\alpha }H{U}_{\alpha }\cup U_{\alpha }H{U}_{-\alpha +1}$$ so that $$\begin{array}{ccc}B{U}_{-\alpha }{n}_{\alpha }n& \subset & B{n}_{\alpha }{U}_{\alpha }{n}_{\alpha }n\cup B{U}_{-\alpha +1}{n}_{\alpha }n\\ & =& B{U}_{-\alpha }n\cup B{n}_{\alpha }n,\text{since}\hspace{0.17em}-\alpha +1\in \left(C\right)\\ & =& Bn{U}_{-w^{-1}\alpha }\cup B{n}_{\alpha }n\\ & \subset & BnB\cup B{n}_{\alpha }nB,\end{array}$$ since $-w^{-1}\alpha \in \left(C\right)\text{.}$ Hence $n_{\alpha }Bn\subset BnB\cup B{n}_{\alpha }nB$ in all cases. $\square$

It follows that all the results of (2.4) are applicable in the present situation. The group $B=P_{\left(C\right)}$ and its conjugates are the Iwahori subgroups of $G\text{;}$ the groups $P_{\left(F\right)}$ and their conjugates, where $F$ is a facet of $C,$ are the parahoric subgroups.

Let $B_0^{+}=Z{U}^{+}\text{.}$ Then $(B_0^{+},N)$ is a BN-pair in $G\text{.}$ We have $B_0^{+}\cap N=Z,$ so that ${\nu }_0$ induces an isomorphism of $N/(B_0^{+}\cap N)$ onto the Weyl group $W_0\text{.}$ The distinguished generators of $N/(B_0^{+}\cap N)$ correspond under this isomorphism to the reflections in the walls of the cone $C_0\text{.}$

		Proof.
	(1) Let $n\in B_0^{+}\cap N\text{.}$ Then $n=z_0{u}^{+}$ $(z_0\in Z,u\in U^{+})$ and therefore $n_0=z_0^{-1}n\in N\cap U^{+}\text{.}$ So we have $n_0=\prod _{\alpha \in {\Sigma }_0^{+}}{u}_a$ with $u_a\in U_a$ by (2.6.3). For each $a\in {\Sigma }_0^{+},$ pick an affine root $\alpha$ with gradient $a$ such that $u_a\in U_{\alpha }\text{.}$ Applying (2.2.1) to this set of affine roots, we see that $n_0\in U_{\left(t{C}_0\right)}$ for some translation $t\in T,$ and hence $n_1=z^{-1}{n}_{0}z\in U_{\left(C_0\right)}$ for some $z\in Z\text{.}$ We have $n_1\in N\cap U_{\left(C_0\right)}\subset N\cap B=H$ by (2.6.8). Hence $n_0\in H\cap U^{+}=\left\{1\right\}$ by axiom (VI). So $n=z_0\in Z$ and therefore $B_0^{+}\cap N=Z\text{.}$ (2) We have to verify the BN-axioms. As before, (BN 1) follows immediately from (I) and (VII). (BN 2): take the distinguished generators of $N/Z$ to be ${\nu }_0^{-1}\left(w_a\right),$ $a\in {\Pi }_0\text{.}$ As to (BN 4), let $a\in {\Pi }_0$ and let $n_a\in {\nu }_0^{-1}\left(w_a\right),$ then if $B_0^{+}=n_a{B}_0^{+}{n}_a^{-1}$ we have $$\begin{array}{ccc}{B}_0=B_0^{+}{n}_a{B}_0^{+}{n}_a^{-1}& =& B_0^{+}{n}_a{U}_a{n}_a^{-1}\text{by (2.6.7)}\\ & =& B_0^{+}{U}_{-a}\end{array}$$ so that $U_{-a}\subset B_0\cap U^{-}=Z{U}^{+}\cap U^{-}=\left\{1\right\}$ by axiom (VI). Since $U_{-a}\ne \left\{1\right\}$ by axiom (II), we have a contradiction. Finally, (BN 3). Let $w\in W_0,$ $a\in {\Pi }_0,$ $n\in {\nu }_0^{-1}\left(w\right),$ $n_a\in {\nu }_0^{-1}\left(w_a\right)\text{.}$ Then $n_a{B}_0^{+}n\subset B_0^{+}{n}_a{B}_0^{+}n=B_0^{+}{n}_a{U}_{a}n$ by (2.6.7). Suppose first that $w^{-1}a$ is positive, then $$B_0^{+}{n}_a{U}_{a}n=B_0^{+}{n}_{a}n{U}_{w^{-1}a}\subset B_0^{+}{n}_{a}n{B}_0^{+}\text{.}$$ If on the other hand $w^{-1}a$ is negative, then $$\begin{array}{ccc}{B}_0^{+}{n}_a{U}_{a}n& =& B_0^{+}{U}_{-a}{n}_{a}n\\ & \subset & B_0^{+}{n}_a{U}_a{n}_{a}n\text{by (2.6.6)}\\ & =& B_0^{+}{U}_{-a}n=B_0^{+}n{U}_{-w^{-1}a}\subset B_0^{+}n{B}_0^{+}\text{.}\end{array}$$ $\square$

Remark. Of course there is a corresponding result for $B_0^{-}=Z{U}^{-},$

$G=Bn{B}_0^{+}\text{.}$

		Proof.
	We remark first that $G$ is generated by $B_0^{+}$ and the $U_{-a}$ $(a\in {\Pi }_0)\text{.}$ For let $G\prime =⟨B_0^{+},{\left(U_{-a}\right)}_{a\in {\Pi }_0}⟩,$ and let $\alpha$ be an affine root with gradient $a\in {\Pi }_0\text{.}$ Then $G\prime$ contains the group $⟨U_{\alpha },U_{-\alpha },H⟩$ and hence by axiom (IV) contains ${\nu }^{-1}\left(w_{\alpha }\right)\text{.}$ Since $G\prime$ contains $Z$ it follows that $N\subset G\prime ,$ hence that $U_{\beta }\subset G\prime$ for all $\beta \in \Sigma \text{.}$ Hence $G\prime =G\text{.}$ Next, let $a\in {\Pi }_0$ and let $C\prime$ be any chamber in $A\text{.}$ Then $$\begin{array}{cc}{U}_{-a}\subset P_{\left(C\prime \right)}N{U}_a\text{.}& \text{(1)}\end{array}$$ For let $\alpha$ be an affine root with gradient $a\text{.}$ If $-\alpha \in \left(C\prime \right)$ we have $U_{-\alpha }\subset P_{\left(C\prime \right)}\text{.}$ If on the other hand $\alpha \in \left(C\prime \right),$ then $$\begin{array}{ccc}{U}_{-\alpha }-U_{-\alpha +1}& \subset & U_{\alpha }·{\nu }^{-1}\left(w_{\alpha }\right)·U_{\alpha }\text{by (2.6.6)}\\ & \subset & P_{\left(C\prime \right)}N{U}_a\text{.}\end{array}$$ This proves (1), from which it follows that $U_{-\alpha }\subset nBN{U}_a$ for all $n\in N,$ so that $$\begin{array}{cc}N{U}_{-a}\subset BN{U}_a\text{.}& \text{(2)}\end{array}$$ Now to show that $G=BN{B}_0^{+},$ it is enough to show that $BN{B}_0^{+}g\subseteq BN{B}_0^{+}$ for all $g$ in a set of generators of $G\text{.}$ Hence it is enough to show that $BN{B}_0^{+}{U}_{-a}\subseteq BN{B}_0^{+}$ for $\alpha \in {\Pi }_0\text{.}$ But $$\begin{array}{ccc}BN{B}_0^{+}{U}_{-a}& =& BN{U}_a{U}^a{U}_{-a}=BN{U}_a{U}_{-a}{U}^a\\ & \subset & BN{U}_a{\nu }_0^{-1}\left(w_a\right)U_a{U}^a\text{by (2.6.6)}\\ & =& BN{\nu }_0^{-1}\left(w_a\right)U_{-a}{U}^{+}\\ & =& BN{U}_{-a}{U}^{+}\\ & \subset & BN{U}_a{U}^{+}\text{by (2) above}\\ & =& BN{U}^{+}=BN{B}_0^{+}\text{.}\end{array}$$ $\square$

Let $T^{+}$ (resp. $T^{++}\text{)}$ be the set of all translations $t\in T$ such that $t\left(0\right)\in {\bar{C}}_0$ (resp. $t\left(0\right)\in C_0^{\perp }\text{).}$ Let $Z^{+}={\nu }^{-1}\left(T^{+}\right),$ $Z^{++}={\nu }^{-1}\left(T^{++}\right),$ so that $Z^{+}$ and $Z^{++}$ are semigroups contained in $Z\text{.}$

Also let $K$ denote the maximal parahoric subgroup $P_{\left(0\right)}\text{.}$

(1)	(Cartan decomposition) $G=K{Z}^{++}K,$ and the mapping $t\mapsto K·{\nu }^{-1}\left(t\right)·K$ is a bijection of $T^{++}$ onto $K\backslash G/K\text{.}$
(2)	(Iwasawa decomposition) $G=KZ{U}^{-},$ and the mapping $t\mapsto K·{\nu }^{-1}\left(t\right)·U^{-}$ is a bijection of $T$ onto $K\backslash G/U^{-}\text{.}$
(3)	If $z_1\in Z^{++}$ and $z_2\in Z$ are such that $K{z}_{1}K\cap K{z}_2{U}^{-}=\varnothing ,$ then $z_1{z}_2^{-1}\in Z^{+}\text{.}$
(4)	If $z\in Z^{++},$ then $KzK\cap Kz{U}^{-}=Kz\text{.}$

		Proof.
	As in (2.4) we shall identify $N/H$ with $W$ by means of the isomorphism induced by $\nu \text{.}$ Thus $KtK$ means $K·{\nu }^{-1}\left(t\right)·K,$ and so on. In particular, $K=B{W}_{0}B$ by (2.3.2). (1) If $w\in W,$ then $KwK=B{W}_{0}w{W}_{0}B$ by (2.6.8) and (2.3.6). Now $W$ is the disjoint union of the double cosets $W_{0}t{W}_0$ as $t$ runs through $T^{++}\text{.}$ Hence $G=BWB$ is the disjoint union of the cosets $KtK$ with $t\in T^{++}\text{.}$ (2) We shall prove (2) with $U^{-}$ replaced by $U^{+}\text{:}$ this merely amounts to replacing the chamber $C_0$ by its opposite, $-C_0\text{.}$ From (2.6.10) we have $$G=BW{B}_0^{+}=B{W}_0{B}_0^{+}\subset B{W}_{0}B·B_0^{+}=K{B}_0^{+}=KZ{U}^{+}\text{.}$$ To show that $K{t}_1{U}^{+}$ and $K{t}_2{U}^{+}$ are disjoint if $t_1\ne t_2,$ it is enough to show that $K\cap B_0^{+}\subset H{U}^{+}\text{.}$ So let $z_0{u}^{+}\in K,$ where $z_0\in Z$ and $u^{+}\in U^{+}\text{.}$ By (2.2.1) there exists $t\in T$ such that $u^{+}\in U_{\left(t{C}_0\right)}=z{U}_{\left(C_0\right)}{z}^{-1}\subset zB{z}^{-1},$ where $z\in {\nu }^{-1}\left(t\right)\text{.}$ Hence $z_0{u}^{+}\in z_{0}zB{z}^{-1}\cap K,$ hence $z_{0}zB$ meets $Kz\text{:}$ or, putting $t_0=v\left(z_0\right)\in T,$ $t_{0}tB$ meets $Kt=B{W}_{0}Bt,$ and hence $t_{0}tB$ meets $BwBt$ for some $w\in W_0\text{.}$ Consequently $B{t}_{0}tB\subset BwBtB\text{.}$ Now by expressing $w\in W_0$ as a reduced word in the generators $w_a$ $(a\in {\Pi }_0),$ it follows easily from the axiom (BN 3) that $BwBtB\subset B{W}_{0}tB\text{.}$ Hence $B{t}_{0}tB\subset B{W}_{0}tB$ and therefore $t_{0}t\in W_{0}t$ by (2.3.1). Hence $t_0\in W_0\cap T,$ so that $t_0=1$ and $z_0\in H,$ as required. The proofs of (3) and (4) are best expressed in the language of buildings (2.4). (3) Suppose that $t_1\in T^{++}$ and $t_2\in T$ are such that $K{t}_{1}K$ meets $K{t}_2{U}^{-}\text{.}$ Then $K{t}_1^{-1}$ contains an element belonging to $u^{-}{t}_2^{-1}K$ for some $u^{-}\in U^{-}\text{.}$ As above, by (2.2.1) we have $u^{-}\in U_{(-t{C}_0)}\subset P_{\left(C\prime \right)}$ for some chamber $C\prime \text{.}$ Hence $K{t}_1^{-1}$ meets $P_{\left(C\prime \right)}{t}_2^{-1}K\text{.}$ Let $x_1=t_1\left(0\right)$ and $C″=t_1\left(C\prime \right)\text{.}$ Then $P_{\left(x_1\right)}=t_{1}K{t}_1^{-1}$ meets $t_1{P}_{\left(C\prime \right)}{t}_2^{-1}K=P_{\left(C″\right)}{t}_1{t}_2^{-1}K\text{.}$ Let $t_0=t_1{t}_2^{-1}$ and put $x_0=t_0\left(0\right)\text{.}$ Let $ℐ$ be the building associated with $(B,N)$ as in (2.4). Identify the affine space $A$ with the apartment ${𝒜}_0$ by means of the bijection $j$ defined in (2.4.1). Let $\rho$ be the retraction of $ℐ$ onto ${𝒜}_0$ with centre $C″$ (2.4.6) and let $g\in P_{\left(x_1\right)}\cap P_{\left(C″\right)}{t}_{0}K\text{.}$ Since $g\in P_{\left(x_1\right)},$ the image under $g$ of the segment $[0,x_1]\subset {𝒜}_0$ is the segment $[g\left(0\right),x_1]\text{.}$ Since $g\in p{t}_{0}K$ for some $p\in P_{\left(C″\right)},$ we have $$g\left(0\right)=p{t}_0\left(0\right)=p\left(x_0\right)\in p{𝒜}_0$$ and therefore $\rho \left(g\left(0\right)\right)=x_0,$ $\rho \left(x_1\right)=x_1$ (because $x_1\in {𝒜}_0\text{).}$ Hence the image of the segment $[g\left(0\right),x_1]$ under the retraction $\rho$ is a polygonal line $\gamma$ in ${𝒜}_0$ joining $x_0$ to $x_1\text{.}$ Consequently there exist real numbers ${\lambda }_i$ with $0={\lambda }_0<{\lambda }_1<\cdots <{\lambda }_n=1$ such that, if $y_i=\rho \left(g\left({\lambda }_i{x}_1\right)\right),$ then $\gamma$ is the union of the segments $[y_{i-1},y_i]$ $(1\le i\le n)\text{.}$ From (2.4.2) and the definition of $\rho$ it follows that there exist elements $w_i\in W$ such that $$[y_{i-1},y_i]=w_i[{\lambda }_{i-1}{x}_1,{\lambda }_i{x}_1]\text{.}$$ Writing $w_i=w_{i,0}{\tau }_i$ with $w_{i,0}\in W_0$ and ${\tau }_i\in T,$ we have $$y_i-y_{i-1}=({\lambda }_i-{\lambda }_{i-1})w_{i,0}\left(x_1\right)\text{.}$$ Now $x_1\in {\bar{C}}_0,$ hence $x_1-w{x}_1\in C_0^{\perp }$ for all $w\in W_0\text{.}$ So we have $$y_i-y_{i-1}=({\lambda }_i-{\lambda }_{i-1})x_1-y_i^{\prime }$$ with $y_i^{\prime }\in C_0^{\perp }\text{.}$ Summing over $i,$ we get $$x_1-x_0=y_n-y_0=x_1-\sum y_i^{\prime }$$ and therefore $x_0=\sum y_i^{\prime }\in C_0^{\perp }\text{.}$ Since $x_0=t_1{t}_2^{-1}\left(0\right),$ we have proved that $t_1{t}_2^{-1}\in T^{+}\text{;}$ as required. (4) Let $t\in T^{++}$ and let $g\in KtK\cap Kt{U}^{-}\text{.}$ Then by (2.2.1) as before we have $g^{-1}\in P_{\left(C_0^{\prime }\right)}{t}^{-1}K\cap K{t}^{-1}K$ for some translate $C_0^{\prime }$ of the cone $-C_0,$ and hence $g_1=g^{-1}t\in P_{\left(C_0^{\prime }\right)}{P}_{\left(x\right)}\cap P_{\left(0\right)}{P}_{\left(x\right)},$ where $x=t^{-1}\left(0\right)\text{.}$ Now $x\in -{\bar{C}}_0,$ and $-{\bar{C}}_0$ intersects $C_0^{\prime }\text{.}$ Hence there exists $y\in C_0^{\prime }$ such that $x$ lies in the segment $\left[0y\right]\text{.}$ Let $g_1\left(x\right)=x\prime \in ℐ\text{.}$ Since $g_1\in p{P}_{\left(x\right)}$ for some $p\in P_{\left(0\right)}=K,$ we have $x\prime =p\left(x\right)$ and $p\left(0\right)=0\text{.}$ Hence $$d(0,x\prime )=d(p\left(0\right),p\left(x\right))=d(0,x)\text{.}$$ Next, since $g_1\in p\prime P_{\left(x\right)}$ for some $p\prime \in P_{\left(C_{\prime }^0\right)},$ we have $x\prime =p\prime \left(x\right)$ and $y=p\prime \left(y\right)\text{.}$ Consequently $$d(x\prime ,y)=d(p\prime \left(x\right),p\prime \left(y\right))=d(x,y)\text{.}$$ It follows that $w\prime \in \left[0y\right]$ by the uniqueness of geodesies in $ℐ$ (2.4.9) and hence that $x\prime =x\text{.}$ Consequently $g_1$ fixes $x,$ i.e. normalizes $P_{\left(x\right)}$ and therefore (2.3.6) $g_1\in P_{\left(x\right)}=t^{-1}Kt\text{.}$ Hence $g\in Kt\text{.}$ $\square$

Groups of $p\text{-adic}$ type

Let $G$ be an 'abstract' group with affine root structure $(N,\nu ,{\left(u_{\alpha }\right)}_{\alpha \in \Sigma })$ of type ${\Sigma }_0\text{.}$ Now suppose in addition that $G$ carries a Hausdorff topology compatible with its group structure such that

(VIII)	$N$ and the $U_{\alpha }$ are closed subgroups of $G\text{;}$
(IX)	$B$ is a compact open subgroup of $G\text{.}$

A Hausdorff topological group carrying an affine root structure satisfying axioms (I)-(IX) inclusive will be called a simply-connected group of $p\text{-adic}$ type. If $𝓀$ is a $p\text{-adic}$ field (i.e. a locally compact field which is neither discrete nor connected) then $𝓀$ is complete with respect to a discrete valuation, and the residue field is finite. The universal Chevalley group $G({\Sigma }_0,𝓀),$ and more generally the group of $𝓀\text{-rational}$ points of a simply-connected simple linear algebraic group defined over $𝓀,$ then inherits a topology from $𝓀$ which satisfies (VIII) and (IX).

The two new axioms have the following consequences:

$H$ is open in $N$ and compact.

		Proof.
	$H=B\cap N$ (2.6.8), so that $H$ is an open subgroup of $N,$ hence closed in $N,$ hence closed in $G\text{.}$ Since $H$ is contained in $B$ it follows that $H$ is compact. $\square$

A closed subset $M$ of $G$ is compact if and only if $M$ meets only finitely
many cosets $BwB\text{.}$

		Proof.
	The cosets $BwB$ are mutually disjoint, open and compact. The assertion follows immediately from this. $\square$

Comparing (2.7.2) with (2.4.13), we see that a subset $M$ of $G$ is bounded in the sense of (2.4) if and only if $M$ has compact closure. Hence, from (2.4.15):

Every compact subgroup of $G$ is contained in a maximal compact
subgroup. The maximal compact subgroups are precisely the maximal parahoric
 subgroups, and form $(l+1)$ conjugacy classes. All parahoric subgroups are
open and compact. $G$ itself is locally compact, but not compact.

The last two assertions follow from the facts that a parahoric subgroup is a finite union of double cosets $BwB,$ and that $G$ itself is not.

For each $\alpha \in \Sigma ,$ the index $(U_{\alpha }:U_{\alpha +1})$ is finite, and $U_{\alpha }$ is compact.

		Proof.
	There exists a chamber $C\prime$ on which $\alpha$ is positive and for which $h_{\alpha }$ is one of the walls. Transforming by an element of $W,$ we may assume that $C\prime =C,$ i.e. that $\alpha \in \Pi \text{.}$ Since $B$ is open and compact, $B/(B\cap w_{\alpha }B{w}_{\alpha })$ is compact and discrete, hence finite. But $$\begin{array}{ccc}B/(B\cap w_{\alpha }B{w}_{\alpha })& \cong & \left(w_{\alpha }B{w}_{\alpha }B\right)/\left(w_{\alpha }B{w}_{\alpha }\right)\\ & =& \left(w_{\alpha }B{w}_{\alpha }{U}_{\alpha }\right)/\left(w_{\alpha }B{w}_{\alpha }\right)\text{by (2.6.7)}\\ & \cong & U_{\alpha }/(U_{\alpha }\cap w_{\alpha }B{w}_{\alpha })\\ & =& U_{\alpha }/U_{\alpha +1}\end{array}$$ because $U_{\alpha }\cap w_{\alpha }B{w}_{\alpha }=U_{\alpha }\cap P_{\left(w_{\alpha }C\right)}=U_{\alpha +1}$ by (2.6.5). Hence $(U_{\alpha }:U_{\alpha +1})$ is finite. Also $U_{\alpha }$ is a closed subgroup of $B,$ hence is compact. $\square$

(1)	For each $a\in {\Sigma }_0,$ the group $U_{\left(a\right)}$ is closed in $G,$ and for each affine root $\alpha$ with gradient $a,$ $U_{\alpha }$ is an open compact subgroup of $U_{\left(a\right)}\text{.}$
(2)	$U^{+}$ is a closed subgroup of $G$ and $U_{\left(C_0\right)}$ is an open compact subgroup of $U^{+}\text{.}$

		Proof.
	(1) By (2.6.5) $U_{\left(a\right)}\cap B$ is of the form $U_{\alpha },$ where $\alpha$ is some affine root with gradient $a\text{.}$ Hence this $U_{\alpha }$ is open in $U_{\left(a\right)}\text{.}$ Conjugating $U_{\alpha }$ by elements of $Z,$ each $U_{\alpha }$ is open in $U_{\left(a\right)}\text{.}$ Since $U_{\alpha }$ is compact (2.7.4) it follows that $U_{\left(a\right)}$ is locally compact and therefore a closed subgroup of $G\text{.}$ (2) $B\cap U^{+}=U_{\left(C_0\right)},$ hence $U_{\left(C_0\right)}$ is open in $U^{+}\text{;}$ also by (2.6.1) $U_{\left(C_0\right)}$ is a product of $U_{\alpha }\text{'s,}$ hence is compact. Hence $U^{+}$ is locally compact, hence closed in $G\text{.}$ $\square$

$B_0^{+}=Z{U}^{+}$ is locally compact.

		Proof.
	$Z$ is locally compact, because it has $H$ as a compact open subgroup by (2.7.1). Also $U^{+}$ is locally compact by (2.7.5), hence so is $B_0^{+}\text{.}$ $\square$

Consider again the group $G={SL}_{l+1}\left(𝓀\right)$ (2.5.1) where $𝓀$ is now a $p\text{-adic}$ field. Let $ℴ$ be the ring of integers, $𝓅$ the maximal ideal of $ℴ$
and $𝓀=ℴ/𝓅$ the (finite) residue field. $ℴ$ is a compact open subring of $𝓀\text{.}$

$G$ is a closed subset of ${𝓀}^{{(l+1)}^2},$ hence is locally compact (and totally disconnected). The maximal compact subgroup $K$ may be taken to be ${SL}_{l+1}\left(ℴ\right)\text{.}$ Let $\bar{G}={SL}_{l+1}\left(k\right),$ then the projection of $ℴ$ onto $k$ induces an epimorphism $\pi :K\to \bar{G}\text{.}$ Let $\bar{B}$ be the group of upper triangular matrices in $\bar{G},$ then $B={\pi }^{-1}\left(\bar{B}\right)$ is an Iwahori subgroup of $G\text{.}$ $B_0^{+}$ is the group of upper triangular matrices in $G,$ and $Z$ is the diagonal matrices.

Notes and references

This is a typed version of the book Spherical Functions on a Group of $p\text{-adic}$ Type by I. G. Macdonald, Magdalen College, University of Oxford. This book is copyright the University of Madras, Madras 5, India and was first published November 1971.

Published by the Ramanujan Institute, University of Madras, and printed at the Baptist Mission Press, 41A Acharyya Jagadish Bose Road, Calcutta 17, India.

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