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

Last update: 13 February 2014

## Chapter III: Spherical functions on a group of $p\text{-adic}$ type

Throughout this chapter and the succeeding ones, $G$ is a simply-connected group of $p\text{-adic}$ type, as defined in (2.7), and the notation and hypotheses of Chapter II remain in force.

### The root system ${\Sigma }_{1}$

For each affine root $\alpha ,$ the group ${U}_{\alpha }$ is of finite index in ${U}_{\alpha -1},$ by (2.7.4). We define $qα= (Uα-1:Uα).$ Then it follows from axiom (I) that $(3.1.1) qwα=qα$ for all $w\in W\text{.}$ In particular, since ${w}_{\alpha +1}\circ {w}_{\alpha }$ sends $\alpha +r$ to $\alpha +r+2$ for all $r\in ℤ,$ we have $(3.1.2) qα=qα+2.$ Hence, if $a\in {\Sigma }_{0}$ and $r\in ℤ,$ ${q}_{a+r}$ is equal to either ${q}_{a}$ or ${q}_{a+1}\text{.}$ But it can happen that ${q}_{a}\ne {q}_{a+1}\text{.}$

As another particular case of (3.1.1) we have $(3.1.3) qa+r=qwa+r$ for all $w\in {W}_{0}\text{.}$

We define a set ${\Sigma }_{1}$ of vectors in ${A}^{*}$ as follows:

 (1) ${\Sigma }_{0}\subseteq {\Sigma }_{1}\subseteq {\Sigma }_{0}\cup \frac{1}{2}{\Sigma }_{0}\text{;}$ (2) If $a\in {\Sigma }_{0},$ then $\frac{1}{2}a\in {\Sigma }_{1}⇔{q}_{a}\ne {q}_{a+1}\text{.}$

${\Sigma }_{1}$ is an irreducible root system in ${A}^{*},$ with Weyl group ${W}_{0}\text{.}$

 Proof. We have to show that ${\Sigma }_{1}$ satisfies the axioms of (2.1), except for (RS 4). (RS 1) and (RS 5) follow immediately from the corresponding facts for ${\Sigma }_{0},$ and (RS 2) is an immediate consequence of (3.1.3). So it remains to show that $b\left({c}^{\vee }\right)\in ℤ$ for all $b,c\in {\Sigma }_{1}\text{.}$ If $b,c$ are both in ${\Sigma }_{0}$ there is nothing to prove. If $c\in \frac{1}{2}{\Sigma }_{0},$ say $c=\frac{1}{2}a,$ then ${c}^{\vee }=2{a}^{\vee },$ and $b\left(2{a}^{\vee }\right)$ is an even integer if $b\in {\Sigma }_{0},$ an integer if $b\in \frac{1}{2}{\Sigma }_{0}\text{.}$ The only remaining possibility is that $c\in {\Sigma }_{0}$ and $b=\frac{1}{2}a\in \frac{1}{2}{\Sigma }_{0}\text{.}$ Then we have to show that $a\left({c}^{\vee }\right)$ is an even integer. Suppose that $a\left({c}^{\vee }\right)$ is odd. The translation ${t}_{c}$ (defined by ${t}_{c}\left(0\right)={c}^{\vee }\text{)}$ is in the affine Weyl group, and ${t}_{c}\left(a\right)=a-a\left({c}^{\vee }\right)\text{.}$ Hence by (3.1.1) we have ${q}_{a}={q}_{a-a\left(c\right)},$ and therefore ${q}_{a}={q}_{a+1}$ by (3.1.2), because $a\left({c}^{\vee }\right)$ is odd. But then $b=\frac{1}{2}a\notin {\Sigma }_{1}\text{.}$ $\square$

For each $a\in {\Sigma }_{0}$ define $qa/2=qa+1 /qa.$ Thus if $b\in {\Sigma }_{0}\cup \frac{1}{2}{\Sigma }_{0}$ we have ${q}_{b}\ne 1⇔b\in {\Sigma }_{1}\text{.}$ From (3.1.3), ${q}_{wb}={q}_{b}$ for all $w\in {W}_{0}\text{.}$

If ${G}_{1},{G}_{2}$ are subgroups of a group and ${G}_{1}\subseteq {G}_{2},$ we write $\left({G}_{1}:{G}_{2}\right)$ to mean ${\left({G}_{2}:{G}_{1}\right)}^{-1}\text{.}$ With this convention,

$(Ua-r:Ua)= qa/2[r/2] qar$ where $\left[\frac{r}{2}\right]$ is the integral part of $\frac{r}{2}\text{.}$

 Proof. The proof is a straightforward calculation, using (3.1.2). $\square$

Let $a\in {\Pi }_{0}\text{.}$ Then from (2.7.4) we have $\left(B:B\cap {w}_{a}B{w}_{a}\right)=\left({U}_{a}:{U}_{a+1}\right)={q}_{a+1}={q}_{a/2}{q}_{a},$ so that $(3.1.6) (BwaB:B)= qa/2qa (a∈Π0).$

For each $w\in W,$ let $q(w)= (BwB:B).$

Let $w={w}_{1}\cdots {w}_{r}$ be a reduced word for $w\in W\text{.}$ Then $q(w)=q(w1) ⋯q(wr).$

 Proof. If $\alpha \in \Pi$ and $w\in W$ are such that $l\left({w}_{\alpha }w\right)>l\left(w\right)$ then (2.3.7) $B{w}_{\alpha }wB=B{w}_{\alpha }B·BwB\text{.}$ Since ${w}_{\alpha }\ne w,$ the cosets $B{w}_{\alpha }B$ and $BwB$ are disjoint, and therefore $q\left({w}_{\alpha }w\right)=q\left({w}_{\alpha }\right)q\left(w\right)\text{.}$ (3.1.7) now follows by induction on the length. $\square$

If $w\in {W}_{0},$ then $q(w)=∏bqb,$ the product being taken over the $b\in {\Sigma }_{1}$ which are positive on ${C}_{0}$ and negative on $w{C}_{0}\text{.}$

 Proof. Let $w={w}_{1}\cdots {w}_{r}$ be a reduced word for $w$ in ${W}_{0},$ where ${w}_{i}={w}_{{a}_{i}},$ ${a}_{i}\in {\Pi }_{0}\text{.}$ Then the roots $a\in {\Sigma }_{0}$ which are positive on ${C}_{0}$ and negative on $w{C}_{0}$ are congruent under ${W}_{0}$ to ${a}_{1},\dots ,{a}_{r}\text{.}$ The result now follows from (3.1.7), (3.1.6) and the fact that ${q}_{wb}={q}_{b}$ for all $b\in {\Sigma }_{1},$ $w\in {W}_{0}\text{.}$ $\square$

For each $b\in {\Sigma }_{1}$ let ${\xi }_{b}$ be an indeterminate over $ℤ$ such that ${\xi }_{wb}={\xi }_{b}$ for all $w\in {W}_{0}\text{.}$ Define $ξ(w)=∏b ξb(w∈W0)$ where as above the product is over the $b\in {\Sigma }_{1}$ which are positive on ${C}_{0}$ and negative on $w{C}_{0},$ and put $Q(ξ)=∑w∈W0 ξ(w).$

$Q\left(\xi \right)$ is the Poincaré polynomial of the root system ${\Sigma }_{1}\text{.}$ It is a polynomial in one variable if all the roots of ${\Sigma }_{1}$ are of the same length (types ${A}_{l},$ ${D}_{l},$ ${E}_{6},$ ${E}_{7},$ ${E}_{8}\text{),}$ two variables if ${\Sigma }_{1}$ is of type ${B}_{l},$ ${C}_{l},$ ${F}_{4}$ or ${G}_{2},$ and three variables in the case of $B{C}_{l}\text{.}$

Since $K=\bigcup _{w\in {W}_{0}}BwB,$ we have $(3.1.9) (K:B)=∑w∈W0 q(w)=Q(q) say.$

### Haar measures

If $\Gamma$ is any locally compact topological group, the modulus function ${\Delta }_{\Gamma }$ (see e.g. [Die1967], Chapter XIV) is a continuous homomorphism of $\Gamma$ into the multiplicative group of positive real numbers, and therefore is equal to 1 on any compact subgroup of $\Gamma \text{.}$

In the case of our group $G,$ it follows (using (2.7.3)) from this remark that the kernel of ${\Delta }_{G}$ contains all parahoric subgroups and hence is the whole of $G\text{:}$ in other words, $G$ is unimodular. So are $K$ and ${U}_{\alpha },$ because they are compact ((2.7.3) and (2.7.4)); so also are ${U}_{\left(a\right)}$ and ${U}^{±},$ because they are unions of compact subgroups. Finally, the group $Z$ is unimodular, for it has $H$ as a compact normal subgroup with $Z/H$ discrete, by (2.7.1).

On the other hand, the groups ${B}_{0}^{±}=Z{U}^{±}$ are not in general unimodular.

We fix Haar measures $dg,$ $dk,$ $dz,$ $d{u}_{a}$ on $G,$ $K,$ $Z,$ ${U}_{\left(a\right)}$ $\left(a\in {\Sigma }_{0}\right)$ respectively, such that $∫Hdg=∫Kdk= ∫Hdz=∫Ua dua=1.$

Next, consider the group ${U}^{+}\text{.}$ Arrange the positive roots in a sequence ${a}_{1},\dots ,{a}_{n}$ such that height $\left({a}_{i}\right)\le$ height $\left({a}_{j}\right)$ if $i Then, as we ought to have proved in (2.6), the group $Uj=∏i≥j U(ai)$ is a normal subgroup of ${U}^{+}$ for $j=1,2,\dots ,n,$ and ${U}_{j}$ is the semi-direct product of ${U}_{\left({a}_{j}\right)}$ and ${U}_{j+1}\text{.}$ Hence if $d{u}_{j+1}$ is a Haar measure on ${U}_{j+1},$ it follows that $d{u}_{j}=d{u}_{{a}_{j}}d{u}_{j+1}$ is a Haar measure on ${U}_{j}\text{.}$ Consequently $(3.2.1) du+=∏i=1n duai,$ where ${u}^{+}=\prod _{i=1}^{n}{u}_{{a}_{i}},$ is a Haar measure on ${U}_{1}={U}^{+},$ for which the subgroup ${U}_{\left({C}_{0}\right)}=\prod _{i=1}^{n}{U}_{{a}_{i}}$ has measure 1.

The left and right Haar measures on ${B}_{0}^{+}=Z{U}^{+}$ are then $dl(zu+) = dzdu+, dr(zu+) = d(zu+z-1) dz=Δ(z)dz du+$ where $\Delta$ is a continuous homomorphism of $Z$ into the multiplicative group ${ℝ}_{+}^{*}$ of positive real numbers, defined by $Δ(z)=d (zu+z-1) /du+.$

For any integrable function on $G$ we have $∫Gf(g)dg= ∫K∫ZU+ f(zku+)dk dr(zu+)$ or symbolically $(3.2.2) dg=Δ(z)dk dzdu+.$

Since $Z$ normalizes each ${U}_{\left(a\right)}$ we have homomorphisms ${\Delta }_{\left(a\right)}:Z\to {ℝ}_{+}^{*}$ defined by $Δa(z)=d (zuaz-1) /dua.$ $\Delta$ and ${\Delta }_{a}$ are trivial on $H,$ hence induce homomorphisms $\delta ,$ ${\delta }_{a}:T\to {ℝ}_{+}^{*}$ such that $\Delta =\delta \circ \nu ,$ ${\Delta }_{a}={\delta }_{a}\circ \nu \text{.}$

From (3.2.1) we have immediately $(3.2.3) Δ(z)= ∏a∈Σ0+ Δa(z).$

Also the value of ${\Delta }_{a}\left(z\right)$ is given by

Let $a\in {\Sigma }_{0},$ $z\in Z,$ $t=\nu \left(z\right)\in T\text{.}$ Then $Δa(z)= δa(t)= qa/212a(x) qaa(x)$ where $x=t\left(0\right)\text{.}$ If $a\left(x\right)\ge 0$ then ${\delta }_{a}\left(t\right)$ is a positive integer.

 Proof. Since $t\left(a\right)=a\circ {t}^{-1}=a-a\left(x\right)$ we have $z{U}_{a}{z}^{-1}={U}_{a-a\left(x\right)}$ and hence $Δa(z) = (zUaz-1:Ua) = (Ua-a(x):Ua) = qa/2[r/2] qar$ by (3.1.5), where $r=a\left(x\right)\text{.}$ Now $x$ belongs to the lattice spanned by the ${b}^{\vee }$ $\left(b\in {\Sigma }_{0}\right)\text{.}$ If ${q}_{a/2}\ne 1,$ that is if $\frac{1}{2}a\in {\Sigma }_{1},$ then $\frac{1}{2}a\left(x\right)\in ℤ$ by (3.1.4) and therefore $r$ is an even integer, i.e. $\left[r/2\right]=\frac{1}{2}a\left(x\right)\text{.}$ $\square$

$Δ(z)=δ(t)= ∏b∈Σ1+ qbb(x).$

 Proof. (3.2.3), (3.2.4). $\square$

If $w\in {W}_{0}$ and $n\in {\nu }^{-1}\left(w\right),$ then $Δwa(nzn-1) =Δa(z), δwa(wtw-1) =δa(t)$ $\text{(}z\in Z,$ $t\in T\text{).}$

For ${q}_{wb}={q}_{b}$ $\left(b\in {\Sigma }_{1}\right)\text{.}$

Let $a\in {\Pi }_{0}\text{.}$ Then $δ(ta)=δa (ta)=qa/2 qa2.$

 Proof. Consider the product $\prod _{b}{\delta }_{b}\left({t}_{a}\right)$ extended over the roots $b\in {\Sigma }_{0}^{+},$ $b\ne a\text{.}$ Since ${w}_{a}$ permutes this set of roots, we have $∏bδb(ta)= ∏bδwab(ta) =∏bδb(ta-1)$ by (3.2.6), so that $\prod _{b}{\delta }_{b}\left({t}_{a}\right)=1\text{.}$ Hence $\delta \left({t}_{a}\right)={\delta }_{a}\left({t}_{a}\right),$ and ${\delta }_{a}\left({t}_{a}\right)={q}_{a/2}{q}_{a}^{2}$ by (3.2.4), since ${t}_{a}\left(0\right)={a}^{\vee }\text{.}$ $\square$

Let $t\in {T}^{++},$ $w\in {W}_{0}\text{.}$ Then $δ(t)12/ δ(wtw-1)12$ is a positive integer.

 Proof. First consider the case where $w={w}_{a}$ $\left(a\in {\Pi }_{0}\right),$ and let $t$ be any element of $T\text{.}$ Put $x=t\left(0\right)\text{.}$ Then $\left({w}_{a}t{w}_{a}\right)\left(0\right)={w}_{a}\left(x\right)=x-a\left(x\right){a}^{\vee }$ so that ${w}_{a}t{w}_{a}=t·{t}_{a}^{-a\left(x\right)}$ and therefore $(3.2.9) δ(t)/δ (watwa) = δ(ta)a(x) = qa/2a(x) qa2a(x) by (3.2.7) = δa(t)2$ Hence $\delta {\left(t\right)}^{\frac{1}{2}}/\delta {\left({w}_{a}t{w}_{a}\right)}^{\frac{1}{2}}$ is a positive integer, by (3.2.4), provided that $a\left(x\right)\ge 0\text{.}$ Now let $t\in {T}^{++}$ and let $w$ be any element of ${W}_{0}\text{.}$ Let $w={w}_{{a}_{r}}\cdots {w}_{{a}_{1}}$ be a reduced word for $w,$ where ${a}_{i}\in {\Pi }_{0}$ $\left(1\le i\le r\right)\text{.}$ Put ${t}_{i}={w}_{{a}_{i}}\cdots {w}_{{a}_{1}}t{w}_{{a}_{1}}\cdots {w}_{{a}_{i}}$ and put ${x}_{i}={t}_{i}\left(0\right)={w}_{{a}_{i}}\cdots {w}_{{a}_{1}}\left(x\right)$ where $x=t\left(0\right)\text{.}$ Then $ai+1(xi)= (wa1⋯waiai+1) (x)≥0$ because ${w}_{{a}_{1}}\cdots {w}_{{a}_{i}}{a}_{i+1}$ is a positive root (it is one of the positive roots which is negative on ${w}^{-1}{C}_{0},$ see (2.1)) and $x\in {\stackrel{‾}{C}}_{0}\text{.}$ Hence from the first part of the proof we have $\delta {\left({t}_{i}\right)}^{\frac{1}{2}}/\delta {\left({t}_{i+1}\right)}^{\frac{1}{2}}\in ℤ$ and therefore $\delta {\left(t\right)}^{\frac{1}{2}}/\delta {\left(wt{w}^{-1}\right)}^{\frac{1}{2}}=\delta {\left({t}_{0}\right)}^{\frac{1}{2}}/\delta {\left({t}_{r}\right)}^{\frac{1}{2}}\in ℤ\text{.}$ $\square$

For the group ${U}^{-}$ we have analogously $(3.2.1') du-=∏i=1n du-ai$ and the right Haar measure on ${B}_{0}^{-}=Z{U}^{-}$ is ${d}_{r}\left(z{u}^{-}\right)=\Delta \prime \left(z\right)dzd{u}^{-},$ where $\Delta \prime \left(z\right)=d\left(z{u}^{-}{z}^{-1}\right)/d{u}^{-}=\prod {\Delta }_{a}\left(z\right)$ over the negative roots. By (3.2.3) and (3.2.4) this is equal to $\Delta {\left(z\right)}^{-1}\text{.}$ Hence $(3.2.10) dr(zu-)=Δ (z)-1dzdu-$ and $(3.2.2') dg=Δ(z)-1 dkdzdu-.$

In the case of a Chevalley group (2.5) all the ${q}_{a}$ are equal to $q,$ the number of elements in the residue field, and all the ${q}_{a/2}$ are equal to 1, so that ${\Sigma }_{1}={\Sigma }_{0}\text{.}$ So in this case we have $δ(t)12= qr(t(0))$ where $r=\frac{1}{2}\sum _{a\in {\Sigma }_{0}^{+}}a\text{.}$

We shall need later the value of the index ${v}_{t}=\left(KtK:K\right),$ where $t\in {T}^{++}\text{.}$ The computation fits in conveniently here.

Let $C\prime$ be any chamber of $\Sigma$ and let $B\prime ={P}_{\left(C\prime \right)}$ be the corresponding Iwahori subgroup. Then $(B′tB′:B′)= δ(t)$ for all $t\in {T}^{++}\text{.}$

 Proof. By (2.6.4) we have $B\prime ={U}_{\left(C\prime \right)}^{-}H{H}_{\left(C\prime \right)}^{+}\text{.}$ If $\alpha =-a+r$ is positive on $C\prime ,$ where $a\in {\Sigma }_{0}^{+},$ then $t\alpha =-a+r+a\left(t\left(0\right)\right)$ is also positive on $C\prime ,$ because $a\left(t\left(0\right)\right)\ge 0\text{.}$ Hence $t{U}_{\left(C\prime \right)}^{-}{t}^{-1}\subset B\prime$ and therefore $B\prime tB\prime =B\prime t{U}_{\left(C\prime \right)}^{+}\text{.}$ Consequently $(B′tB′:B′) = (B′tU(C′)+t-1:B′) = (tU(C′)+t-1:B′∩tU(C′)+t-1) = (tU(C′)+t-1:U(C′)+) by (2.6.5) = δ(t).$ $\square$

Let $t\in {T}^{++}\text{.}$ Then $q(wtw-1)=δ(t)$ for all $w\in {W}_{0}\text{.}$

 Proof. $q(wtw-1)= (Bwtw-1B:B) =(B′tB′:B′)$ where $B\prime ={w}^{-1}Bw\text{.}$ $\square$

Let $t\in {T}^{++}$ and let ${W}_{0t}$ be the centralizer of $t$ in ${W}_{0},$ or equivalently the subgroup of ${W}_{0}$ which fixes the point $t\left(0\right)\in A\text{.}$ Put $(3.2.14) Qt(ξ)= ∑w∈W0tξ(w)$ in the notation of (3.1), and let $Qt(ξ)=Q(ξ) /Qt(ξ)$ which is also a polynomial.

${v}_{t}=\left(KtK:K\right)=\delta \left(t\right){Q}^{t}\left({q}^{-1}\right)$ for all $t\in {T}^{++}\text{.}$

Proof.

By (2.3.5) we have $KtK=B{W}_{0}t{W}_{0}B,$ which is the disjoint union of the sets $B{t}_{1}{W}_{0}B$ as ${t}_{1}$ runs through the set $\left\{wt{w}^{-1}:w\in {W}_{0}\right\}\text{.}$ For each such ${t}_{1}$ there exists a unique ${w}_{1}\in {W}_{0}$ such that ${t}_{1}^{-1}C\subset {w}_{1}{C}_{0},$ i.e., $C\subset {t}_{1}{w}_{1}{C}_{0}\text{.}$ Thus we have $(Bt1W0B:B) = ∑w∈W0 (Bt1w1wB:B) = ∑w∈W0 q(t1w1w).$ Now by (2.2.2) and (3.1.7) we have $q\left({t}_{1}{w}_{1}w\right)=q\left({t}_{1}{w}_{1}\right)·q\left(w\right)\text{.}$ In particular, taking $w={w}_{1}^{-1},$ it follows that $q(t1w1)=q (t1)·q(w1)-1$ and therefore $(Bt1W0B:B)= ∑w∈W0q(t1) ·q(w1)-1q (w)$ or, making use of (3.2.12), $(3.2.16) (Bt1W0B:B)= δ(t)q(w1)-1 Q(q).$

Now the subgroup ${W}_{0t}$ of ${W}_{0}$ has a unique set ${W}_{0}^{t}$ of left coset representatives with the property that $wt∈W0tand wt∈W0t⇒l (wtwt)=l (wt)+l(wt).$

Let ${w}_{0}$ be the unique element of ${W}_{0}$ of maximum length, and let ${w}_{0t}$ be the unique element of maximum length in ${W}_{0t}\text{.}$ Then, with ${w}_{1}$ as above, we have

${w}_{1}{w}_{0}{w}_{0t}\in {W}_{0}^{t}\text{.}$

 Proof. We have ${t}_{1}=wt{w}^{-1}$ and we may assume that $w\in {W}_{0}^{t}\text{.}$ Since $w{t}^{-1}{w}^{-1}C\subset {w}_{1}{C}_{0},$ we have $w{t}^{-1}\left(0\right)\in {w}_{1}{\stackrel{‾}{C}}_{0},$ hence ${w}_{1}^{-1}w{t}^{-1}\left(0\right)\in {\stackrel{‾}{C}}_{0}\text{;}$ but also ${w}_{0}{t}^{-1}\left(0\right)\in {\stackrel{‾}{C}}_{0},$ because $t\in {T}^{++}\text{.}$ Hence ${w}_{1}^{-1}w$ and ${w}_{0}$ both map ${t}^{-1}\left(0\right)$ to the same point, because ${\stackrel{‾}{C}}_{0}$ is a fundamental domain for ${W}_{0}\text{.}$ It follows that ${w}_{0}{w}_{1}^{-1}w$ belongs to ${W}_{0t}\text{.}$ We want to show that it is equal to ${w}_{0t},$ i.e. that it changes the sign of every $a\in {\Sigma }_{0}^{+}$ such that $a\left(t\left(0\right)\right)=0\text{.}$ So let $a>0,$ $a\left(t\left(0\right)\right)=0$ and let $y\in C\text{.}$ Then $z={w}_{1}^{-1}{t}_{1}^{-1}y\in {C}_{0}\text{.}$ We have $w1-1wa(z)= a(t-1w-1y) = a(w-1y-t(0)) = wa(y)because a(t(0))=0$ and since $w\in {W}_{0}^{t}$ we have $wa>0,$ hence $wa\left(y\right)>0\text{.}$ Hence ${w}_{1}^{-1}wa>0$ and therefore ${w}_{0}{w}_{1}^{-1}wa<0,$ as required. $\square$

So we have $w1w0=wt w0t$ for some ${w}^{t}\in {W}_{0}^{t}\text{.}$ Since $l(wtw0t) = l(wt)+ l(w0t), l(w1w0) = l(w0)- l(w1)$ it follows that $q(wt) q(w0t)= q(w0) q(w1)-1$ and therefore $(3.2.18) ∑q(w1)-1 = q(w0t)q(w0) ∑q(wt) = Qt(q-1).$

Hence finally $(KtK:K) = (K:B)-1 ∑w∈W0t (Bwtw-1W0B:B) = Q(q)-1∑δ(t) q(w1)-1Q(q) by (3.2.16) = δ(t)Qt(q-1) by (3.2.18).$

$\square$

### Zonal spherical functions on $G$ relative to $K$

Let $ℒ\left(G,K\right)$ be the algebra of $K\text{-bi-invariant}$ functions on $G$ defined in (1.1), and let $ℋ\left(G,K\right)$ be the subring consisting of the $ℤ\text{-valued}$ functions in $ℒ\left(G,K\right)\text{.}$ $ℋ\left(G,K\right)$ is called the Hecke ring of $G$ relative to $K\text{.}$ Clearly $ℒ\left(G,K\right)\cong ℋ\left(G,K\right){\otimes }_{ℤ}ℂ\text{.}$

For each $t\in {T}^{++}$ let ${\chi }_{t}$ be the characteristic function of the double coset $KtK\text{.}$ By (1.1.1) and (2.6.11) the ${\chi }_{t}$ form a $ℤ\text{-basis}$ of $ℋ\left(G,K\right)$ and a $ℂ\text{-basis}$ of $ℒ\left(G,K\right)\text{.}$ The identity element is ${\chi }_{1},$ the characteristic function of $K\text{.}$

Let $s:T\to {ℂ}^{*}$ be a homomorphism. Then $\left(s{\delta }^{\frac{1}{2}}\right)\circ \nu$ is a continuous homomorphism of $Z$ into ${ℂ}^{*},$ which we extend trivially across ${U}^{-}$ to a continuous homomorphism ${\varphi }_{s}:{B}_{0}^{-}\to {ℂ}^{*}\text{:}$ $ϕs(zu-)= (sδ12,ν(z)).$ [To avoid large collections of brackets, we shall sometimes write $\left(s,t\right)$ for $s\left(t\right),$ where $s\in \text{Hom}\left(T,{ℂ}^{*}\right)$ and $t\in T\text{.]}$ Now extend ${\varphi }_{s}$ to a function on the whole of $G$ by the rule $ϕs(kzu-)= (sδ12,ν(z))$ $\text{(}k\in K,$ $z\in Z,$ ${u}^{-}\in {U}^{-}\text{).}$

For each $s\in \text{Hom}\left(T,{ℂ}^{*}\right)$ the function $ωs(g)=∫K ϕs(g-1k) dk$ is a zonal spherical function on $G$ relative to $K\text{.}$ Moreover, if $s$ is a character of $T,$ i.e. if $|s\left(t\right)|=1$ for all $t\in T,$ then ${\omega }_{s}$ is positive definite.

 Proof. This is a direct consequence of the results of Chapter I. Since $K\cap {B}_{0}^{-}\subset H{U}^{-}$ (see the proof of (2.6.11)(2)) it follows that ${\varphi }_{s}$ restricted to ${B}_{0}^{-}$ is trivial on $K\cap {B}_{0}^{-}$ and hence (1.2.7) is a zonal spherical function on ${B}_{0}^{-}$ relative to $K\cap {B}_{0}^{-}\text{.}$ Hence ${\omega }_{s}$ is a z.s.f. on $G$ relative to $K,$ by (1.3.2). The second assertion follows from (1.4.7) and (3.2.10). $\square$

For all ${g}_{0}\in G$ and $s\in \text{Hom}\left(T,{ℂ}^{*}\right),$ $ωs-1(g0) =ωs(g0-1).$

 Proof. Let $\stackrel{‾}{K}=K/\left(K\cap {B}_{0}^{-}\right)$ and let $d\stackrel{‾}{k}$ be a relatively invariant measure on $\stackrel{‾}{K}$ with total mass $1\text{.}$ If ${g}_{0}^{-1}k=k\prime z{u}^{-}$ $\text{(}k\prime \in K,$ $z\in Z,$ ${u}^{-}\in {U}^{-}\text{)}$ then as in (1.4.6) we have $d\stackrel{‾}{k}=\Delta {\left(z\right)}^{-1}d\stackrel{‾}{k}\prime ,$ and hence $ωs-1(g0) = ∫K‾ ϕs-1 (g0-1k) dk‾ = ∫K‾ (s-1δ12,ν(z)) Δ(z)-1d k‾′ = ∫K‾(sδ12,ν(z-1)) dk‾′ = ∫K‾ϕs (g0k′)dk‾ ′=ωs(g0-1).$ $\square$

The group ${W}_{0}$ acts on $T$ by inner automorphisms, hence acts on $\text{Hom}\left(T,{ℂ}^{*}\right)\text{:}$ namely $(ws,t)= (s,w-1tw)$ $\text{(}s\in \text{Hom}\left(T,{ℂ}^{*}\right),$ $t\in T,$ $w\in {W}_{0}\text{).}$

For all $w\in {W}_{0}$ and $s\in \text{Hom}\left(T,{ℂ}^{*}\right),$ $ωs=ωws.$

This result is a consequence of the calculations we shall undertake in Chapter IV, and we shall postpone its proof to (4.5). The reader can check that there is no vicious circle involved here.

Now let $f\in ℒ\left(G,K\right)\text{.}$ Then if ${\stackrel{ˆ}{\omega }}_{s}$ is the spherical measure associated with ${\omega }_{s},$ $ωˆs(f) = (f*ωs)(1) = ∫Gf(g) ωs(g-1) dg = ∫Gf(g) (∫Kϕs(gk)dk) dg = ∫Gf(g)ϕs (g)dg = ∫K∫Z∫U-f (kzu-) (sδ12,ν(z)) Δ(z)-1 dkdzdu- = ∫Z(sδ-12,ν(z)) (∫U-f(zu-)du-) dz.$

The integral $\Delta {\left(z\right)}^{-\frac{1}{2}}{\int }_{{U}^{-}}f\left(z{u}^{-}\right)d{u}^{-}$ depends only on $f$ and $\nu \left(z\right)=t$ say. We shall therefore write $f∼(t)=δ (t)-12 ∫U-f(zu-) du-$ for $t\in T$ and any $z\in {\nu }^{-1}\left(t\right)\text{.}$ Then the calculation above shows that $(3.3.4) ωˆs(f)= ∑t∈Tf∼ (t)s(t).$

The $\stackrel{\sim }{f}\left(t\right)$ should be regarded as the Fourier coefficients of $f\in ℒ\left(G,K\right)\text{.}$ There are no convergence difficulties here: since $f$ has compact support and the double cosets $Kt{U}^{-}$ are open (because $K$ is open) and mutually disjoint, it follows that $f$ takes non-zero values on only finitely many of these double cosets, i.e. $\stackrel{\sim }{f}$ has finite support in $T\text{.}$

Let $ℂ\left[T\right]$ be the group algebra of $T$ over $ℂ\text{.}$ Since $T$ is a free abelian group generated by the ${t}_{a}$ with $a\in {\Pi }_{0},$ it follows that $ℂ\left[T\right]=ℂ\left[{t}_{a}^{±1}:a\in {\Pi }_{0}\right],$ hence is a commutative integral domain. Consequently $ℂ\left[T\right]$ is the coordinate ring (or affine algebra) of an affine algebraic variety, say $S,$ whose points are the $ℂ\text{-algebra}$ homomorphisms $s:ℂ\left[T\right]\to ℂ\text{.}$ Restricting these homomorphisms to $T$ gives a bisection of $S$ onto $\text{Hom}\left(T,{ℂ}^{*}\right),$ and we shall identify $\text{Hom}\left(T,{ℂ}^{*}\right)$ with $S$ in this way. The elements of $ℂ\left[T\right]$ can therefore be regarded as functions on $S=\text{Hom}\left(T,{ℂ}^{*}\right)\text{.}$ Hence, by the Nullstellensatz, if $\varphi \in ℂ\left[T\right],$ $(3.3.5) ϕ=0⇔s(ϕ)=0 for all s∈S.$

We regard the affine Weyl group $W$ as a discrete group. Then, in the notation of (1.1), $ℒ\left(T\right)$ is the algebra of functions on $T$ with finite support. The mapping $ϕ⟼∑t∈T ϕ(t)t$ is a $ℂ\text{-algebra}$ isomorphism of $ℒ\left(T\right)$ onto $ℂ\left[T\right]\text{.}$ Hence $s\in S$ defines a homomorphism $ℒ\left(T\right)\to ℂ,$ namely $s(ϕ)=∑t∈T ϕ(t)s(t).$ The formula (3.3.4) now takes the form $(3.3.4') ωˆs(f)= s(f∼).$

The group ${W}_{0}$ acts on $ℒ\left(T\right)$ by transposition: $\left(w\varphi \right)\left(t\right)=\varphi \left({w}^{-1}tw\right)$ $\text{(}\varphi \in ℒ\left(T\right),$ $t\in T,$ $w\in {W}_{0}\text{).}$ Let ${ℒ\left(T\right)}^{{w}_{0}}$ denote the subring of ${W}_{0}\text{-invariant}$ functions in $ℒ\left(T\right)\text{.}$ Then we have

The mapping $f↦\stackrel{\sim }{f}$ is a $ℂ\text{-algebra}$ isomorphism of $ℒ\left(G,K\right)$ onto ${ℒ\left(T\right)}^{{w}_{0}}\text{.}$

$ℒ\left(G,K\right)$ is commutative.

 Proof. First of all, let us see that $\stackrel{\sim }{f}\in ℒ{\left(T\right)}^{{W}_{0}}$ whenever $f\in ℒ\left(G,K\right)\text{.}$ Let $s\in S,$ $w\in {W}_{0}\text{.}$ Then $s(f∼)=ωˆs (f)=ωˆws (f)=(ws) (f∼)=s(w-1f∼)$ by (3.3.3) and (3.3.4'). Hence $\stackrel{\sim }{f}={w}^{-1}\stackrel{\sim }{f}$ by (3.3.5). Next, let ${f}_{1},{f}_{2}\in ℒ\left(G,K\right)\text{.}$ Then $s(f1*f2)∼ =ωˆs (f1*f2)= ωˆs(f1)· ωˆs(d2) = s(f∼1)· s(f∼2) = s(f∼1*f∼2)$ by (3.3.4'). Again by (3.3.5) we deduce that ${\left({f}_{1}*{f}_{2}\right)}^{\sim }={\stackrel{\sim }{f}}_{1}*{\stackrel{\sim }{f}}_{2},$ and hence $f↦\stackrel{\sim }{f}$ is a $ℂ\text{-algebra}$ homomorphism. It remains to show that the mapping $f↦\stackrel{\sim }{f}$ of $ℒ\left(G,K\right)$ into $ℒ{\left(T\right)}^{{W}_{0}}$ is bijective. Since the characteristic functions ${\chi }_{t}$ $\left(t\in {T}^{++}\right)$ form a $ℂ\text{-basis}$ of $ℒ\left(G,K\right),$ it is enough to show that the ${\stackrel{\sim }{\chi }}_{t}$ form a $ℂ\text{-basis}$ of $ℒ{\left(T\right)}^{{W}_{0}}\text{.}$ The last two parts of (2.6.11) were designed expressly for this purpose. For each $t\in T,$ the orbit of $t$ under ${W}_{0}$ (acting by inner automorphisms) has a unique representative in ${T}^{++}\text{.}$ Let $⟨t⟩$ be the characteristic function of the ${W}_{0}\text{-orbit}$ of $t\text{.}$ Then these functions $⟨t⟩,$ as $t$ runs through ${T}^{++},$ form a $ℂ\text{-basis}$ of $ℒ{\left(T\right)}^{{W}_{0}}\text{.}$ Hence we can write $(3.3.8) χ∼t= ∑t′ χ∼t (t′) ⟨t′⟩$ summed over $t\prime \in {T}^{++}\text{.}$ We define a total ordering on $T$ as follows. Let ${\Pi }_{0}=\left\{{a}_{1},\dots ,{a}_{l}\right\}$ be the set of simple roots of ${\Sigma }_{0},$ arranged in some order. For each $t\in T$ we have $t\left(0\right)=\sum _{i=1}^{l}{m}_{i}{a}_{i}^{\vee }$ with ${m}_{i}\in ℤ\text{.}$ We define $(3.3.9) { t≥0⇔ the first non-zero mi is positive, t1≥t2⇔ t1t2-1 ≥0.$ In particular, the elements of ${T}^{+}$ are $\ge 0$ for this ordering. Now if $t\in {T}^{++}$ and $t\prime \in T$ we have $∫Kt′U- χt(g)dg= ∫U-χt (z′u-) Δ(z′)-1 du-$ for any $z\prime \in {\nu }^{-1}\left(t\prime \right),$ $=δ(t′)-12 χ∼t(t′).$ Hence we have $(3.3.10) χ∼t(t′)= δ(t′)12 vol.(KtK∩Kt′U-).$ So if ${\stackrel{\sim }{\chi }}_{t}\left(t\prime \right)\ne 0$ it follows from (2.6.11)(3) that $t{t\prime }^{-1}\in {T}^{+}$ and hence that $t\ge t\prime \text{.}$ Again, from (2.6.11)(4) we have ${\stackrel{\sim }{\chi }}_{t}\left(t\right)=\delta {\left(t\right)}^{\frac{1}{2}}\text{.}$ Consequently (3.3.8) now takes the form $(3.3.8') χ∼t=δ (t)12 ⟨t⟩+ ∑t′ from which it follows immediately (since $\delta \left(t\right)$ is never $0\text{)}$ that the ${\stackrel{\sim }{\chi }}_{t}$ form a basis of $ℒ{\left(T\right)}^{{W}_{0}}\text{.}$ $\square$

(a) Consider the algebra $ℒ\left(W,{W}_{0}\right)\text{.}$ Restriction of functions to the subgroup $T$ defines an isomorphism $\rho :ℒ\left(W,{W}_{0}\right)\to ℒ{\left(T\right)}^{{W}_{0}}\text{.}$ Hence (3.3.6) provides a canonical isomorphism of $ℒ\left(G,K\right)$ onto $ℒ\left(W,{W}_{0}\right),$ namely $f↦{\rho }^{-1}\left(\stackrel{\sim }{f}\right)\text{.}$ (The emphasis here is on the word 'canonical'.)

(b) If $G$ is a universal Chevalley group (2.5) over a $p\text{-adic}$ field $𝓀,$ there is a much simpler proof that $ℒ\left(G,K\right)$ is commutative. For there exists an  nvolutory antiautomorphism $i:G\to G$ such that $i\left({u}_{a}\left(\xi \right)\right)={u}_{-a}\left(\xi \right)$ for all $a\in {\Sigma }_{0}$ and $\xi \in 𝓀,$ and $i\left(z\right)=z$ for all $z\in Z\text{.}$ It follows that $i\left(K\right)=K$ and that $i$ preserves Haar measure. Now let $f\in ℒ\left(G,K\right)$ and define ${f}^{i}=f\circ i\text{.}$ Since $i$  is an antiautomorphism, we have ${\left({f}_{1}*{f}_{2}\right)}^{i}={f}_{2}^{i}*{f}_{1}^{i}\text{.}$ On the other hand, $i$ preserves all double cosets $KzK$ and therefore $f={f}^{i}$ for all $f\in ℒ\left(G,K\right)\text{.}$ Hence ${f}_{1}*{f}_{2}={f}_{2}*{f}_{1}\text{.}$

The same argument will work whenever $G$ has an antiautomorphism $i$ such that $i\left({U}_{a+r}\right)={U}_{-a+r}$ for all $a\in {\Sigma }_{0}$ and $r\in ℤ,$ and $i\left(z\right)\in zH$ for all $z\in Z\text{.}$ This will be the case, for example, whenever the central inversion ${w}_{0}:x↦-x$ is in the Weyl group ${W}_{0}\text{.}$ For then we may take $i\left(g\right)={n}_{0}{g}^{-1}{n}_{0}^{-1},$ where ${n}_{0}\in {\nu }^{-1}\left({w}_{0}\right)\text{.}$

(c)From (3.3.8') it follows that the image of $ℋ\left(G,K\right)$ in $ℒ{\left(T\right)}^{{W}_{0}}$ under  $f↦\stackrel{\sim }{f}$ is the $ℤ\text{-module}$ spanned by the elements $\delta {\left(t\right)}^{\frac{1}{2}}⟨t⟩$ with $t\in {T}^{++}$ (because $\delta {\left(t\prime \right)}^{-\frac{1}{2}}{\stackrel{\sim }{\chi }}_{t}\left(t\prime \right)\in ℤ$ by (3.3.10)). If ${\Sigma }_{0}={\Sigma }_{1},$ then $\delta {\left(t\right)}^{\frac{1}{2}}\in ℤ$ by (3.2.7), hence in this case the isomorphism of $ℒ\left(G,K\right)$ onto $ℒ\left(W,{W}_{0}\right)$ in (a) above embeds $ℋ\left(G,K\right)$ in $ℋ\left(W,{W}_{0}\right),$ the Hecke ring of $W$ relative to ${W}_{0}\text{.}$

 (1) Every zonal spherical function $\omega$ on $G$ relative to $K$ is equal to ${\omega }_{s}$ for some $s\in S=\text{Hom}\left(T,{ℂ}^{*}\right)\text{.}$ (2) ${\omega }_{s}={\omega }_{s\prime }⇔s\prime =ws$ for some $w\in {W}_{0}\text{.}$

 Proof. (1) The mapping $\stackrel{\sim }{f}↦\stackrel{ˆ}{\omega }\left(f\right)$ is a $ℂ\text{-algebra}$ homomorphism $ℒ{\left(T\right)}^{{W}_{0}}\to ℂ,$ which since $ℒ\left(T\right)$ is integral over $ℒ{\left(T\right)}^{{W}_{0}}$ can be extended to a homomorphism $ℒ\left(T\right)\to ℂ\text{.}$ This in turn defines by restriction a homomorphism $s:T\to {ℂ}^{*},$ and we have $\stackrel{ˆ}{\omega }\left(f\right)=s\left(\stackrel{\sim }{f}\right)={\stackrel{ˆ}{\omega }}_{s}\left(f\right)$ for $f\in ℒ\left(G,K\right)\text{.}$ Hence $\stackrel{ˆ}{\omega }={\stackrel{ˆ}{\omega }}_{s}$ and therefore $\omega ={\omega }_{s}$ by (1.7). (2) Suppose that ${\omega }_{s}={\omega }_{s\prime }\text{.}$ Then $s$ and $s\prime$ agree on $ℒ{\left(T\right)}^{{W}_{0}},$ by (3.3.4'), and therefore their kernels $m,$ $m\prime ,$ say, are maximal ideals of $ℒ\left(T\right)$ lying over the same maximal ideal of $ℒ{\left(T\right)}^{{W}_{0}}\text{.}$ By a well-known theorem of commutative algebra (see for example [AMa1969], Chapter V, Ex. 13) there exists $w\in {W}_{0}$ transforming $m$ into $m\prime \text{.}$ Hence $s\prime =ws\text{.}$ $\square$

Since ${ℂ}^{*}\cong ℝ×\left(ℝ/ℤ\right)$ (the isomorphism being $r{e}^{2\pi i\theta }↦\left(\text{log} r,\theta \right)\text{)}$ we have $S=\text{Hom}\left(T,{ℂ}^{*}\right)\cong \text{Hom}\left(T,ℝ\right)×\text{Hom}\left(T,ℝ/ℤ\right)\text{.}$ Now $\text{Hom}\left(T,ℝ\right)$ can be identified with ${A}^{*}\text{:}$ if $\lambda \in \text{Hom}\left(T,ℝ\right),$ the corresponding element of ${A}^{*}$ is the linear form which maps $t\left(0\right)$ to $\lambda \left(t\right)$ for all $t\in T\text{.}$ This identification induces an identification of $\text{Hom}\left(T,ℝ/ℤ\right)$ with ${A}^{*}/L,$ where (as in (2.5)) $L$ is the lattice of linear forms $u$ on $A$ such that $u\left({a}^{\vee }\right)\in ℤ$ for all $a\in {\Sigma }_{0}\text{.}$ Hence the space $\Omega$ of z.s.f. may be identified with the orbit space $\left({A}^{*}×\left({A}^{*}/L\right)\right)/{W}_{0}\text{.}$ Alternatively, $\Omega$ may be identified with $S/{W}_{0},$ the affine algebraic variety whose coordinate ring is $ℒ{\left(T\right)}^{{W}_{0}}\cong ℂ{\left[T\right]}^{{W}_{0}}\text{.}$

Let ${t}_{1},\dots ,{t}_{m}$ be a set of generators of the semigroup ${T}^{++}\text{.}$ Then $ℋ\left(G,K\right)=ℤ\left[{\chi }_{{t}_{1}},\dots ,{\chi }_{{t}_{m}}\right]\text{.}$ (Of course, in general the ${\chi }_{{t}_{i}}$ will not be algebraically independent.)

 Proof. For each $t\in {T}^{++}$ let $θt=δ (t)12 ⟨t⟩$ so that the image of $ℋ\left(G,K\right)$ under $f↦\stackrel{\sim }{f}$ is the $ℤ\text{-module}$ freely generated by the ${\theta }_{t},$ by (3.3.11)(c). Let ${M}_{t}=\sum _{t\prime From (3.3.8') we have ${\stackrel{\sim }{\chi }}_{t}\equiv {\theta }_{t} \left(\text{mod} {M}_{t}\right)\text{.}$ Also $θt1* θt2≡ θt1t2 (mod Mt1t2).$ For ${\theta }_{{t}_{1}}*{\theta }_{{t}_{2}}-{\theta }_{{t}_{1}{t}_{2}}$ is a sum of terms of the form $δ(t1t2)12 δ(t3)-12 θts$ with ${t}_{3}={t}_{1}^{\prime }{t}_{2}^{\prime }\in {T}^{++}$ and ${t}_{1}^{\prime }={w}_{1}{t}_{1}{w}_{1}^{-1},$ ${t}_{2}^{\prime }={w}_{2}{t}_{2}{w}_{2}^{-1}$ $\text{(}{w}_{1},{w}_{2}\in {W}_{0}\text{)}$ and either ${t}_{1}^{\prime }\ne {t}_{1}$ or ${t}_{2}^{\prime }\ne {t}_{2}\text{.}$ By (3.2.8) the coefficient $\delta {\left({t}_{1}{t}_{2}\right)}^{\frac{1}{2}}\delta {\left({t}_{3}\right)}^{-\frac{1}{2}}$ is a rational integer, and we have ${t}_{3}<{t}_{1}{t}_{2}\text{.}$ Hence if $t=\prod _{i=1}^{m}{t}_{i}^{{\mu }_{i}}\in {T}^{++}$ it follows that $χ∼t≡θt≡ ∏i=1m θtiμi≡ ∏i=1m χ∼tiμi (mod Mt)$ and so, by induction on $t,$ that ${\stackrel{\sim }{\chi }}_{t}\in ℤ\left[{\stackrel{\sim }{\chi }}_{{t}_{1}},\dots ,{\stackrel{\sim }{\chi }}_{{t}_{m}}\right]\text{.}$ Hence ${\chi }_{t}\in ℤ\left[{\chi }_{{t}_{1}},\dots ,{\chi }_{{t}_{m}}\right]\text{.}$ $\square$

## 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.