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

Last update: 19 February 2014

## Chapter IV: Calculation of the spherical functions

### Statement of result

For each $b\in {\Sigma }_{0}\cup \frac{1}{2}{\Sigma }_{0}$ let ${t}_{b}\in T$ be the translation defined by ${t}_{b}\left(0\right)={b}^{\vee }$ (so that ${t}_{a/2}={t}_{a}^{2}\text{).}$ Let $s\in \text{Hom}\left(T,{ℂ}^{*}\right)$ and define $c(b,s)= 1-qb/2-1/2 qb-1s (tb)-1 1-qb/2-1/2 s(tb)-1$ whenever the denominator is non-zero. (If $b\notin {\Sigma }_{1},$ then ${q}_{b}$ is taken to be $1\text{.}$ Likewise for ${q}_{b/2}\text{.)}$ If $a\in {\Sigma }_{0}$ we put $c0(a,s)= c(12a,s) c(a,s).$ So if $\frac{1}{2}a\notin {\Sigma }_{1}$ we have $c0(a,s)= c(a,s)= 1-qa-1 s(ta)-1 1-s(ta)-1$ whilst if $\frac{1}{2}a\in {\Sigma }_{1}$ we have $c0(a,s) = 1-qa/2-1 s(ta)-2 1-s(ta)-2 · 1-qa/2-1/2 qa-1s (ta)-1 1-qa/2-1/2 s(ta)-1 = ( 1+qa/2-1/2 s(ta)-1 ) ( 1-qa/s-1/2 qa-1s(ta)-1 ) 1-s(ta)-2 .$ Finally put $c(s)= ∏a∈Σ1+ c(b,s)= ∏a∈Σ0+ c0(a,s).$ Then $c\left(s\right)$ is defined whenever $s\left({t}_{b}\right)\ne 1$ for all $b\in {\Sigma }_{1}\text{.}$ The function $c$ is the counterpart of Harish-Chandra's $c\text{-function}$ for a real semisimple Lie group.

 (1) $c\left(b,s\right)=c\left(wb,ws\right)$ for $b\in {\Sigma }_{1}$ and $w\in {W}_{0}\text{.}$ (2) ${c}_{0}\left(a,s\right)+{c}_{0}\left(-a,s\right)=1+q{\left({w}_{a}\right)}^{-1}\text{.}$ (3) $c\left({s}^{-1}\right)=c\left({w}_{0}s\right),$ where ${w}_{0}$ is the element of ${W}_{0}$ which sends every positive root to a negative root. Proof. (1) is clear since ${q}_{wb}={q}_{b}$ for all $w\in {W}_{0}\text{.}$ (2) is easily checked, remembering that $q\left({w}_{a}\right)={q}_{a/2}{q}_{a}$ (3.1.6). (3) follows from (1) and the definitions. $\square$

The main result of this chapter is the following:

Let $s:T\to {ℂ}^{*}$ be a homomorphism such that $s\left({t}_{b}\right)\ne 1$ for all $b\in {\Sigma }_{1}\text{.}$ Then if ${z}_{0}\in {Z}^{++}$ we have $ωs(z0-1)= δ(t0)-1/2 Q(q-1) ∑w∈W0c (ws)(ws,t0)$ where ${t}_{0}=\nu \left({z}_{0}\right)$ and $Q\left({q}^{-1}\right)=\sum _{w\in {W}_{0}}q{\left(w\right)}^{-1}\text{.}$

The proof will occupy sections (4.2)-(4.5). To see what the formula looks like in a particular case, take $G={SL}_{l+1}\left(𝓀\right)$ as in (2.7.7). Here ${\Sigma }_{1}={\Sigma }_{0}$ and ${q}_{a}=q$ for all roots $a,$ where $q$ is the number of elements in the residue field of $𝓀\text{.}$ The roots are ${e}_{i}-{e}_{j}$ $\left(i\ne j\right)$ in the notation of (2.2.2) and we may write $s\left({t}_{{e}_{i}-{e}_{j}}\right)={\xi }_{i}{\xi }_{j}^{-1},$ where the ${\xi }_{i}$ are non-zero complex numbers, determined up to a common factor by $s\text{.}$ Then $c(s)=∏i Let $\pi$ be a generator of the maximal ideal, then we may take ${z}_{0}$ to be the diagonal matrix $z0= ( πλ0 ⋱ πλl )$ with ${\lambda }_{0}\ge {\lambda }_{1}\ge \dots \ge {\lambda }_{l}$ (and $\sum {\lambda }_{i}=0\text{).}$ Then $\left(s,{t}_{0}\right)={\xi }_{0}^{{\lambda }_{0}}\cdots {\xi }_{l}^{{\lambda }_{l}}\text{;}$ also by (3.2.11) $\delta {\left(t\right)}^{1/2}={q}^{n\left(\lambda \right)}$ where $n\left(\lambda \right)=l{\lambda }_{0}+\left(l-1\right){\lambda }_{1}+\dots +{\lambda }_{l-1},$ and $Q(q-1)= ∏i-1l 1-q-(i+1) 1-q-1 .$ So we have $ωs(z-1)= q-n(λ) Q(q-1) ∑σ ξσ(0)λ0⋯ ξσ(l)λi ∏i summed over all permutations $\sigma$ of $\left\{0,1,\dots ,l\right\}\text{.}$

### Preliminary reductions

From (3.3.1) we have $(4.2.1) ωs(g)= ∫Kϕs (g-1k) dk$ where ${\varphi }_{s}\left(kz{u}^{-}\right)=\left(s{\delta }^{1/2},\nu \left(z\right)\right)\text{.}$ We shall prove (4.1.2) by computing this integral explicitly.

First we remark that ${\varphi }_{s}$ is constant on each double coset $Kt{U}^{-}\text{.}$ For fixed $g\in G,$ the set ${g}^{-1}K$ meets only finitely many of these double cosets (because they are open and mutually disjoint, and ${g}^{-1}K$ is compact), and therefore the integrand in (4.2.1) takes only finitely many values, each of which is of the form $\left(s{\delta }^{1/2},t\right)$ for some $t\in T\text{.}$ Hence ${\omega }_{s}\left(g\right)$ is of the form $s\left({\Phi }_{g}\right)$ for some ${\Phi }_{g}\in ℂ\left[T\right]$ independent of $s\text{.}$

Next, since ${\omega }_{s}$ is constant on each double coset $KtK$ we may assume that $g\in {Z}^{++}\text{.}$ Write $g={z}^{-1}$ and $t=\nu \left(z\right),$ so that ${t}^{-1}\in {T}^{++}\text{.}$ Then $(4.2.2) ωs(z-1) = ∫Kϕs (zk)dk = ∑w∈W0 ∫BwB ϕs(zk) dk.$

Consider the integral ${\int }_{BwB}{\varphi }_{s}\left(zk\right)dk\text{.}$ We have to look at ${\varphi }_{s}\left(g\right)$ as $g$ runs through $zBwB=z{U}_{\left({C}_{0}\right)}wB={U}_{t\left({C}_{0}\right)}twB\text{.}$ Now ${t}^{-1}\left(0\right)\in {\stackrel{‾}{C}}_{0},$ hence $0\in \stackrel{‾}{t\left({C}_{0}\right)}$ and therefore ${U}_{t\left({C}_{0}\right)}\subset K\text{.}$ Hence the integration is effectively over the set $twB,$ or equivalently (since $w\in K\text{)}$ over $t\prime B$ where $t\prime ={w}^{-1}tw\text{.}$

Next, we have $B={U}_{\left({C}_{0}\right)}H{U}_{\left(C\right)}^{-}$ by (2.6.4), so that the integration is effectively over the set $z\prime {U}_{\left({C}_{0}\right)},$ where $z\prime$ is any element of ${\nu }^{-1}\left(t\prime \right)\text{;}$ and $z\prime {U}_{\left({C}_{0}\right)}={U}_{\left({C}_{0}^{\prime }\right)}z\prime$ where ${C}_{0}^{\prime }=t\prime \left({C}_{0}\right)$ is a translate of the cone ${C}_{0}\text{.}$ So we have $(4.2.3) ∫BwBϕs (zk)dk=κ ∫U(C0′) ϕs(u+z′) du+$ where $d{u}^{+}$ is the Haar measure on ${U}^{+},$ normalized as in (3.2), and $\kappa$ is a constant independent of $s\text{.}$ To evaluate $\kappa$ we may put $s={\delta }^{-1/2},$ so that ${\varphi }_{s}$ is identically $1\text{.}$ Then the left-hand side of (4.2.3) is equal to the volume of $BwB,$ which is $q\left(w\right)/Q\left(q\right)$ in the notation introduced in (3.1), and the right-hand side is equal to $κ(U(C0′:U(C0))) =κ·δ(t′),$ so that $κ=q(w)Q(q) δ(t′)-1.$ But also ${\varphi }_{s}\left({u}^{+}z\prime \right)={\varphi }_{s}\left({u}^{+}\right)·\left(s{\delta }^{1/2},t\prime \right),$ and therefore (4.2.3) becomes $(4.2.4) ∫BwBϕs (zk)dk= q(w)Q(q) (sδ-1/2,t′) ∫U(C0′) ϕs(u+)d u+.$

So if we put $(4.2.5) Jw(s)= (w(sδ-1/2),t) ∫U(C0′) ϕs(u+)du+$ where ${C}_{0}^{\prime }=t\prime \left({C}_{0}\right)={w}^{-1}tw\left({C}_{0}\right),$ then from (4.2.2) and (4.2.4) we have $(4.2.6) ωs(z-1)= 1Q(q) ∑w∈W0q(w) Jw(s).$

Since ${U}_{\left({C}_{0}^{\prime }\right)}$ is compact it intersects only finitely many of the double cosets $Kt{U}^{-},$ and so as before we conclude that $(4.2.7) Jw(s)=s (Φz,w) for some Φz,w ∈ℒ(T).$

To evaluate ${\omega }_{s}\left({z}^{-1}\right)$ we shall compute the integrals ${J}_{w}\left(s\right)\text{.}$ First of all, if $w=1,$ then $t\prime =t$ and ${\stackrel{‾}{C}}_{0}^{\prime }=t\left({\stackrel{‾}{C}}_{0}\right)$ contains the origin $0,$ so that ${U}_{\left({C}_{0}^{\prime }\right)}\subset K$ and therefore ${\varphi }_{s}\left({u}^{+}\right)=1$ for all ${u}^{+}\in {U}_{\left({C}_{0}^{\prime }\right)}\text{.}$ Consequently $(4.2.8) J1(s)= (sδ1/2,t)$ (because the volume of ${U}_{\left({C}_{0}^{\prime }\right)}$ is $\delta \left(t\right)\text{).}$

Now assume that $w\ne 1\text{.}$ Then there exists a simple root $a\in {\Pi }_{0}$ such that $wa\in {\Sigma }_{0}^{-}\text{.}$ The group ${U}_{\left({C}_{0}\right)}$ is the semi-direct product of ${U}_{a}$ and the group ${U}^{a}=⟨{U}_{b}:b\in {\Sigma }_{0}^{+},b\ne a⟩\text{.}$ Hence ${U}_{\left({C}_{0}^{\prime }\right)}=z\prime {U}_{\left({C}_{0}\right)}{z\prime }^{-1}$ is the semi-direct product of the group $V=z′Uaz′-1 ⊂U(a)$ where $U(a)= ⟨U(b):b∈Σ0+,b≠a⟩,$ and the group $z′Uaz′-1= Ut′(a).$ Hence $(4.2.9) Jw(s)= (w(sδ-1/2),t) ∫V∫Ut′(a) ϕs(uaua)d uadua$ where $d{u}^{a},$ $d{u}_{a}$ are the Haar measures on ${U}^{\left(a\right)}$ and ${U}_{\left(a\right)}$ respectively, normalized as in (3.2).

In order to compute ${J}_{w}\left(s\right)$ we write ${U}_{t\prime \left(a\right)}$ as the difference between the sets ${U}_{\left(a\right)}$ and ${U}_{\left(a\right)}-{U}_{t\prime \left(a\right)},$ and consider the integrals $(4.2.10) { Jw1(s)= (w(sδ-1/2),t) ∫V∫U(a) ϕs(uaua) duadua, Jw0(s)= (w(sδ-1/2),t) ∫V∫U(a)-Ut′(a) ϕs(uaua) duadua.$

Here the integration is no longer over a compact set, since neither ${U}_{\left(a\right)}$ nor ${U}_{\left(a\right)}-{U}_{t\prime \left(a\right)}$ is compact, and we shall therefore have to restrict $s$ in order that the integrals shall converge.

### The integral ${\int }_{{U}_{\left(a\right)}}{\varphi }_{s}\left({u}_{a}\right)d{u}_{a}$

Let $a\in {\Sigma }_{0}^{+}$ and let $s:T\to {ℂ}^{*}$ be such that $|s\left({t}_{a}\right)|>1\text{.}$ Then the integral $∫U(a) ϕs(ua) dua$ converges, and its value is equal to ${c}_{0}\left(a,s\right)=c\left(\frac{1}{2}a,s\right)c\left(a,s\right)$ in the notation of (4.1). Proof. We split up ${U}_{\left(a\right)}$ into the union of ${U}_{a}$ and the subsets $Yr=Ua-r- Ua-r+1 (r≥1).$ If ${u}_{a}\in {U}_{a},$ then ${u}_{a}\in K$ and therefore ${\varphi }_{s}\left({u}_{a}\right)=1\text{.}$ Consequently $∫Uaϕa (ua)dua=1.$ Next consider ${u}_{a}\in {Y}_{r}$ $\left(r\ge 1\right)\text{.}$ From (2.6.6) we can write $ua=u1nu2$ where ${u}_{a},{u}_{2}\in {U}_{-a+r}$ and $\nu \left(n\right)={w}_{a-r}\text{.}$ Hence $ϕs(ua)= ϕs(u1nu2) =ϕs(n)$ because ${U}_{-a+r}\subset {U}^{-}\cap K\text{.}$ Choosing ${n}_{a}\in N$ such that $\nu \left({n}_{a}\right)={w}_{a}$ we have ${n}_{a}\in K$ and therefore $ϕs(ua)= ϕs(nan)= (sδ1/2,wa∘wa-r) =(sδ1/2,ta-r).$ Hence $∫Yrϕs( ua)dua= (sδ1/2,ta-r) ·volume of Yr$ and by (3.1.5) the volume of ${Y}_{r}$ is equal to $qa/2[r/2] qar- qa/2[(r-1)/2] qar-1.$ Consequently the value of the integral is $1+∑r=1∞s (ta)-r qa/2-r/2 qa-r ( qa/2[r/2] qa- qa/2[(r-1)/2] qar-1 ) ,$ making use of (3.2.7). This is a geometric series with common ratio $s{\left({t}_{a}\right)}^{-2},$ hence converges if and only if $|s\left({t}_{a}\right)|>1\text{.}$ Summing the series, we get the value announced in (4.3.1). $\square$

### Reduction of ${J}_{w}\left(s\right)$

Consider first the integral ${J}_{w}^{1}\left(s\right)$ defined in (4.2.10). We recall that in the definition of ${J}_{w}\left(s\right),$ $a$ is a simple root such that $wa\in {\Sigma }_{0}^{-}\text{.}$

Suppose that $|s\left({t}_{a}\right)|>1\text{.}$ Then the integral ${J}_{w}^{1}\left(s\right)$ converges and its value is $c0(a,s) Jw′(was)$ where $w\prime =w{w}_{a}\text{.}$ Proof.

Choose ${n}_{a}\in N$ such that $\nu \left({n}_{a}\right)={w}_{a}\text{.}$ Since ${n}_{a}\in K$ we have ${\varphi }_{s}\left({u}^{a}{u}_{a}\right)={\varphi }_{s}\left({n}_{a}^{-1}{u}^{a}{n}_{a}·{n}_{a}^{-1}{u}_{a}\right)\text{.}$ Write ${n}_{a}^{-1}{u}^{a}{n}_{a}$ in the form $k{z}_{1}{u}^{-},$ where $k\in K,$ ${z}_{1}\in Z,$ ${u}^{-}\in {U}^{-}\text{.}$ Then $ϕs(uaua) = ϕs ( z1u-· na-1ua ) = ϕs ( z2·na u-na-1 ·ua )$ where ${z}_{2}={n}_{a}{z}_{1}{n}_{a}^{-1},$ and ${n}_{a}{u}^{-}{n}_{a}^{-1}\in {n}_{a}{U}^{-}{n}_{a}^{-1}\in {U}^{\left(-a\right)}·{U}_{\left(a\right)}\text{.}$ Hence ${n}_{a}{u}^{-}{n}_{a}^{-1}$ can be written uniquely in the form ${u}^{-a}{u}_{a}^{\prime }$ $\text{(}{u}^{-a}\in {U}^{\left(-a\right)},$ ${u}_{a}^{\prime }\in {U}_{\left(a\right)}\text{)}$ and therefore $ϕs(uaua)= ϕs(z2u-aua′ua)= ϕs(z2ua′ua)$ since ${U}_{\left(a\right)}$ normalizes ${U}^{\left(-a\right)}$ and ${\varphi }_{s}$ is invariant on the right with respect to ${U}^{-}\supset {U}^{\left(-a\right)}\text{.}$ So, putting ${u}_{a}^{″}={u}_{a}^{\prime }{u}_{a},$ we have $ϕs(uaua)= ϕs(z2ua″)= ϕs(z2ua″z2-1) (sδ1/2,ν(z2))$ and $(sδ1/2,ν(z2)) = (was,ν(z1)) (δ1/2,ν(z2)) = ϕwas (na-1uana)· (δ1/2,ν(z1-1z2)).$ Hence $ϕs(uaua)= ϕs(z2ua″z2-1)· ϕwas (na-1uana)· (δ1/2,ν(z1-1z2)).$ As ${u}^{a}$ runs through the subgroup $V=z\prime {U}^{a}{z\prime }^{-1}$ defined in (4.2), ${n}_{a}^{-1}{u}^{a}{n}_{a}$ runs through $V\prime ={n}_{a}^{-1}V{n}_{a}$ and so we have $Jw1(s)= (w(sδ-1/2),t) ∫V′ϕwas (ua) (δ1/2,ν(z1-1z2)) dua×∫U(a) ϕs(z2ua″z2-1) dua.$ But $∫U(a)ϕs (z2ua″z2-1) dua = ∫U(a)ϕs (z2uaz2-1) dua = Δa(z2)-1 ∫U(a)ϕs (ua)dua$ and by (4.3.1) this integral converges, since $|s\left({t}_{a}\right)|>1\text{.}$ Since $V\prime$ is compact it follows that ${J}_{w}^{1}\left(s\right)$ converges, and we have $(4.4.2) Jw1(s)= (w(sδ-1/2),t) c0(a,s) ∫V′ϕwas (ua)dua$ since by (3.2.9) $(δ1/2,ν(z1-1z2))= δa(ν(z2))= Δa(z2).$

To complete the proof of (4.4.1) we shall show that

For any $s\in \text{Hom}\left(T,{ℂ}^{*}\right),$ $Jw′(s)= (w′s,t) (wδ-1/2,t) ∫V′ϕs (ua)dua.$ Proof. From the definition (4.2.5), $Jw′(s)= (w′(sδ-1/2),t) ∫U(C0″) ϕs(ua)dua$ where ${C}_{0}^{″}=\left({w\prime }^{-1}tw\prime \right)\left({C}_{0}\right),$ so that ${U}_{\left({C}_{0}^{″}\right)}$ is the semi-direct product of $V\prime$ and ${U}_{\beta },$ where $\beta =\left({w\prime }^{-1}tw\prime \right)a=a-w\prime a\left(x\right)$ where $x=t\left(0\right)\text{.}$ Since $w\prime =w{w}_{a},$ we have $β(0)=a(0)+ wa(x)=wa(x)$ which is $\ge 0$ because $wa$ is a negative root, and $x\in -{\stackrel{‾}{C}}_{0}\text{.}$ Hence $U\subset K,$ so that $∫Uβϕs(ua) dua = volume of Uβ = δa(w′-1tw′) = δ(w′-1tw′)1/2/ δ(w-1tw)1/2 by (3.2.9) = (w′δ1/2,t)· (wδ-1/2,t).$ This completes the proof of (4.4.3) and hence also of (4.4.1). $\square$

$\square$

Next we consider the integral $Jw0(s)= (w(sδ-1/2),t) ∫V∫U(a)-Uα ϕs(uaua)d uadua$ where $\alpha =\left({w}^{-1}tw\right)\left(a\right)=a-wa\left(x\right)$ if $x=t\left(0\right)\text{.}$

Suppose that $|s\left({t}_{a}\right)|>1\text{.}$ Then the integral ${J}_{w}^{0}\left(s\right)$ converges and is equal to $(c0(a,s)-1) Jw′(s)$ where as before $w\prime =w{w}_{a}\text{.}$ Proof. As in (4.3) put $Yr=Ua-r-Ua-r+1$ so that ${U}_{\left(a\right)}-{U}_{\alpha }$ is the disjoint union of the sets ${Y}_{r}$ for $r\ge 1+wa\left(x\right)>0\text{.}$ Consider the integral $∫V∫Yrϕs (uaua)dua dua.$ The calculation proceeds on similar lines to those of (4.3). By (2.6.6) we may write ${u}_{a}={u}_{1}n{u}_{2}$ with ${u}_{1},{u}_{2}\in {U}_{-a+r}$ and $\nu \left(n\right)={w}_{a-r}\text{.}$ Hence $ϕs(uaua) = ϕs(uau1n) (because u2∈U-) = ϕs(u1-1uau1na-1·nan) (because u1∈K)$ where ${n}_{a}\in {\nu }^{-1}\left({w}_{a}\right),$ $=ϕs (u1-1uau1na-1) ·(sδ1/2,ta-r)$ because ${n}_{a}n\in ℤ$ and $\nu \left({n}_{a}n\right)={w}_{a}{w}_{a-r}={t}_{a}^{-r}\text{.}$ Hence $∫V∫Yrϕs (uaua)dua dua= ( ∫Vϕs (u1-1uau1na-1) dua ) ×(sδ1/2,ta-r) vol. Yr.$ But ${u}_{1}$ normalizes $V$ and does not change the measure, so that $∫Vϕs (u1-1uau1na-1) dua = ∫Vϕs (uana-1) dua = ∫V′ϕs (ua)dua$ where as before $V\prime ={n}_{a}^{-1}V{n}_{a}\text{.}$ Consequently $∫V∫U(a)-Uα ϕs(uaua)d uadua= (∫V′ϕs(ua)dua) ∑r=1+wa(x)∞ (sδ1/2,ta)-r ( qa/2[r/2] qar- qa/2[(r-1)/2] qar-1 ) .$ Summing this geometric series, which converges if and only if $|s\left({t}_{a}\right)|>1,$ and using (4.4.3) now leads to the result. [Recall that $wa\left(x\right)$ is an even integer if ${q}_{a/2}\ne 1,$ by (3.1.4).] $\square$

These calculations lead to the following reduction formula:

If $s\left({t}_{a}\right)\ne 1,$ then $Jw(s)=c0 (a,s)Jw′ (was)- (c0(a,s)-1) Jw′(s),$ where $a\in {\Pi }_{0}$ is such that $wa$ is negative, and $w\prime =w{w}_{a}\text{.}$ Proof. If $|s\left({t}_{a}\right)|>1,$ this follows directly from (4.4.1) and (4.4.4). By (4.2.7) the $J\text{'s}$ are polynomials in ${\xi }_{i}=s\left({t}_{{a}_{i}}\right)$ and ${\xi }_{i}^{-1}$ $\left(1\le i\le l\right),$ where ${a}_{1},\dots ,{a}_{l}$ are the simple roots. Hence it is clear (e.g. by analytic continuation) that (4.4.5) is valid whenever the right-hand side is defined, i.e. whenever $s\left({t}_{a}\right)\ne 1\text{.}$ $\square$

This reduction formula is the key result. In the next section we shall use it first to prove that ${\omega }_{s}={\omega }_{ws}$ for all $w\in {W}_{0},$ which was postponed from Chapter III (3.3.3), and then to complete the proof of the formula (4.1.2).

### End of the proof

We shall begin by proving (3.3.3). Clearly it is enough to show that ${\omega }_{s}\left({z}^{-1}\right)={\omega }_{{w}_{a}s}\left({z}^{-1}\right)$ for all $a\in {\Pi }_{0},$ and we may assume that $s\left({t}_{a}\right)\ne 1,$ otherwise $s={w}_{a}s$ and there is nothing to prove.

Put ${I}_{w}\left(s\right)={J}_{w}\left(s\right)-{J}_{w}\left({w}_{a}s\right)$ and suppose first that $wa$ is a negative root, and $a\in {\Pi }_{0}\text{.}$ Then by (4.4.5) $Jw(s)=-c0 (a,s)Iw′ (s)+Jw′(s)$ where $w\prime =w{w}_{a}\text{.}$ Replacing $s$ by ${w}_{a}s$ and subtracting, we get $Iw(s) = ( 1-c0(a,s) -c0(-a,s) ) Iw′(s) = -q(wa)-1 Iw′(s)$ by (4.1.1). Since $wa$ is negative, there is a reduced word for $w$ ending with  ${w}_{a},$ and therefore $q\left(w\right)=q\left(w\prime \right)q\left({w}_{a}\right)$ by (3.1.7). Hence we have  $(4.5.1) q(w)Iw(s)=- q(w′)Iw′ (s).$ If on the other hand $wa$ is positive, then the roles of $w\prime$ and $w$ are interchanged, and (4.5.1) is therefore true for all $w\in {W}_{0}\text{.}$ Summing over $w\in {W}_{0},$ it follows that $∑w∈W0q(w) Iw(s)=0$ and hence ${\omega }_{s}\left({z}^{-1}\right)={\omega }_{{w}_{a}s}\left({z}^{-1}\right)$ by (4.2.6).

Let $w\in {W}_{0}$ and let $w={w}_{1}\cdots {w}_{r}$ be a reduced word in the generators ${w}_{a},$ $a\in {\Pi }_{0}\text{.}$ Let $E$ be the set of all products ${w}_{1}^{{\epsilon }_{1}}\cdots {w}_{r}^{{\epsilon }_{r}}$ where ${\epsilon }_{i}$ is $0$ or $1\text{.}$

If $s\left({t}_{b}\right)\ne 1$ for all $b\in {\Sigma }_{1},$ we shall say that $s$ is nonsingular.

Suppose $s$ is nonsingular. Then $Jw(s)=δ (t)1/2 ∑w′∈E λw,w′(s) ·(w′s,t)$ with coefficients ${\lambda }_{w,w\prime }\left(s\right)$ independent of $t\text{.}$ Also $λw,w(s)= ∏ac0 (a,s)$ the product being taken over the roots $a\in {\Sigma }_{0}^{+}$ such that $wa$ is negative. Proof. by induction on the length of $l\left(w\right)\text{.}$ If $l\left(w\right)=0$ then $w=1$ and ${J}_{1}\left(s\right)=\delta {\left(t\right)}^{1/2}\left(s,t\right)$ by (4.2.8). If $l\left(w\right)>0,$ apply (4.4.5) with $w\prime ={w}_{1}\cdots {w}_{r-1}\text{.}$ $\square$

From (4.5.2) and (4.2.6) it follows that we have $(4.5.3) {\omega }_{s}(z-1)= δ(t)1/2 ∑w∈W0 μw(s) (ws,t)$ with coefficients ${\mu }_{w}\left(s\right)$ independent of $t=\nu \left(z\right)\text{.}$ Consider in particular the coefficient ${\mu }_{{w}_{0}}\left(s\right),$ where ${w}_{0}$ is the unique element of ${W}_{0}$ which sends every positive root to a negative root. This coefficient comes only from ${J}_{{w}_{0}}\left(s\right)$ and hence by (4.5.2) we have $(4.5.4) μw0(s)= q(w0) Q(q) λw0,w0 (s)= c(s) Q(q-1)$ since $q{\left({w}_{0}\right)}^{-1}Q\left(q\right)=Q\left({q}^{-1}\right)\text{.}$

Next we have

Assume that $|s|$ is nonsingular (i.e. that $|s\left({t}_{a}\right)|\ne 1$ for all $a\in {\Sigma }_{0}\text{).}$ Then the homomorphisms $ws$ with $w\in {W}_{0}$ are linearly independent over $ℂ\text{.}$

Assuming this for the moment, since ${\omega }_{s}={\omega }_{w\prime s}$ for all $w\prime \in {W}_{0}$ we have from (4.5.3) $∑w∈W0μw (s)(ws,t)= ∑w∈W0μw (w′s)(ww′s,t)$ and therefore (assuming $|s|$ nonsingular) by (4.5.5) ${\mu }_{w}\left(s\right)={\mu }_{1}\left(ws\right)$ for all $w\in {W}_{0}\text{.}$ In particular, therefore $μ1(w0s)= c(s)Q(q-1)$ by (4.5.4); or, replacing $s$ by ${w}_{0}s,$ $μ1(s)= c(s-1)Q(q-1)$ by (4.1.1). Hence (4.5.3) now becomes $(4.5.6) ωs(z-1)= δ(t)1/2 Q(q-1) ∑w∈W0c (ws-1) (ws,t)$ provided that $|s|$ is nonsingular. However, this restriction is superfluous, for the left-hand side of (4.5.6) is of the form $s\left({\Phi }_{t}\right)$ for some ${\Phi }_{t}\in ℂ\left[T\right],$ and the right-hand side is of the form $s\left({\Psi }_{t}\right)$ for some ${\Psi }_{t}\in ℂ\left(T\right),$ the field of fractions of $ℂ\left[T\right]\text{.}$ Since ${\Phi }_{t}$ and ${\Psi }_{t}$ take the same values on a dense open set in $S=\text{Hom}\left(T,{ℂ}^{*}\right),$ it follows that they are equal and therefore (4.5.6) is valid whenever the right-hand side is defined, i.e. provided $s$ is nonsingular.

Finally, to derive the formula (4.3.2) we have merely to replace $s$ by ${s}^{-1}$ and put ${z}_{0}={z}^{-1}$ in (4.5.6), bearing in mind (3.3.2).

There remains the proof of (4.5.5). This will follow from

Let ${s}_{1},\dots ,{s}_{n}$ $\left(n\ge 1\right)$ be distinct elements of $\text{Hom}\left(T,{ℂ}^{*}\right)\text{.}$ Then they are linearly independent. Proof. The argument is a familiar one. If the ${s}_{i}$ are linearly dependent, choose a linear relation between them containing as few non-zero terms as possible, say $∑i=1rλi si=0$ (renumbering the ${s}_{i}$ if necessary) with each ${\lambda }_{i}\ne 0\text{.}$ Since ${s}_{1}\ne {s}_{r}$ there exists ${t}_{0}\in T$ such that ${s}_{1}\left({t}_{0}\right)\ne {s}_{r}\left({t}_{0}\right)\text{.}$ Then we have $∑i=1rλi si(t)=0$ and $∑i=1rλisi (t0)si(t)=0$ for all $t\in T,$ so by subtraction we get $∑i=1rλi (si(t0)-sr(t0)) si=0$ contradicting the minimality of $r\text{.}$ $\square$

From (4.5.7) it follows that if the set $\left\{ws:w\in {W}_{0}\right\}$ is linearly dependent, then ${w}_{1}s={w}_{2}s$ for some ${w}_{1}\ne {w}_{2}$ in ${W}_{0},$ and hence $s$ is fixed by an element $\ne 1$ in ${W}_{0}\text{.}$ Hence the same is true of $\text{log} |s|\in \text{Hom}\left(T,ℝ\right)\text{.}$ Identifying $\text{log} |s|$ with an element $u\in {A}^{*}$ as in (3.3.13), it follows that $u$ has non-trivial stabilizer under the action of ${W}_{0}$ on ${A}^{*}$ and hence lies on a reflecting hyperplane. In other words, we have $u\left({a}^{\vee }\right)=0$ for some $a\in {\Sigma }_{0},$ and this means that $|s\left({t}_{a}\right)|=1\text{.}$ This proves (4.5.5) and hence completes the proof of (4.1.2).

If $s={\delta }^{-1/2}$ (which is nonsingular) then ${\omega }_{s}$ is identically $1\text{.}$ Hence also ${\omega }_{{\delta }^{1/2}}=1$ by (3.3.2). Putting $s={\delta }^{1/2}$ in (4.1.2) gives therefore $Q(q-1)=δ (t0)-1/2 ∑w∈W0c (wδ1/2) (wδ1/2,t0)$ for all ${t}_{0}\in {T}^{++}\text{.}$ Consider $c\left(w{\delta }^{1/2}\right)$ for $w\ne 1\text{.}$ There exists a simple root $a$ such that ${w}^{-1}a$ is negative. Let $b=-{w}^{-1}a,$ then we have ${t}_{b}^{-1}=w{t}_{a}{w}^{-1}$ and therefore $wδ1/2 (tb)-1= δ1/2(ta)= qa/21/2 qa=qb/21/2 qb$ so that $c\left(b,w{\delta }^{1/2}\right)=0$ and therefore $c\left(w{\delta }^{1/2}\right)=0\text{.}$ Consequently the sum above reduces to the term corresponding to $w=1,$ that is $Q(q-1) = c(δ12) = ∏b∈Σ1+ 1-qb/2-1/2 qb-1δ (tb)-1/2 1-qb/2-1/2 δ(tb)-1/2 .$ This suggests (but does not prove) the following multiplicative formula for the Poincare polynomial $Q\left(\xi \right)$ of the root system ${\Sigma }_{1}\text{.}$ For $b\in {\Sigma }_{1},$ define $ηb=∏c∈Σ1+ ξcc(b∨).$ Then $(4.5.9) Q(ξ)= ∏b∈Σ1+ 1-ξb/21/2 ξbηb1/2 1-ξb/21/2 ηb1/2$ with the convention that ${\xi }_{b/2}=1$ if $\frac{b}{2}\notin {\Sigma }_{1}\text{.}$ In fact (4.5.9) is true and can be proved by elementary methods. Putting ${\xi }_{b}=\xi$ for all $b\in {\Sigma }_{1},$ and assuming that ${\Sigma }_{1}$ is reduced (i.e. that ${\Sigma }_{1}={\Sigma }_{0}\text{),}$ (4.5.9) specializes to give $∑w∈W0 ξl(w) = ∏a∈Σ0+ 1-ξ1+ht(a∨) 1-ξht(a∨) = ∏a∈Σ0+ 1-ξ1+ht(a) 1-ξht(a)$ (since ${\Sigma }_{0}$ and ${\Sigma }_{0}^{\vee }$ have the same Weyl group ${W}_{0}\text{).}$

### The singular case

Next we consider the problem of finding an expression for ${\omega }_{s}\left({z}^{-1}\right)$ when $s$ is singular, i.e. when $s\left({t}_{b}\right)=1$ for some $b\in {\Sigma }_{1}\text{.}$ We have remarked earlier that ${\omega }_{s}\left({z}^{-1}\right)$ is of the form $s\left({\Psi }_{t}\right)$ (where $t=\nu \left(z\right)\text{)}$ for some polynomial ${\Psi }_{t}\in ℂ{\left[T\right]}^{{W}_{0}},$ independent of $s\text{.}$ Hence we can calculate ${\omega }_{s}\left({z}^{-1}\right)$ for singular $s$ as the limit of ${\omega }_{{s}_{1}}\left({z}^{-1}\right)$ as ${s}_{1}\to s$ through nonsingular values of ${s}_{1}\text{.}$ This limit can be computed by applying L'Hopital's rule in exactly the same way that the degree of an irreducible representation of a compact Lie group is derived from Weyl's character formula.

For simplicity we shall consider first the 'most singular' case, namely $s=1\text{.}$ Then we have to evaluate the limit as $s\to 1$ of $∑w∈W0c (ws)(ws,t) (t∈T++).$ In calculating the limit, we may make $s\to 1$ through real values. So let $s$ be real and let $\sigma$ be the linear form on $A$ defined by $\sigma \left(t\left(0\right)\right)=\text{log} s\left(t\right)$ for all $t\in T\text{.}$ Then $s\left({t}_{a}\right)={e}^{\sigma \left({a}^{\vee }\right)},$ so that $c(s)=∏a∈Σ0+ (1+qa/2-1/2e-σ(a∨)) (1-qa/2-1/2qa-1e-σ(a∨)) 1-e-σ(2a∨) .$ Let us temporarily write $∏a∈Σ0+ (1+qa/2-1/2e-σ(a∨)) (1-qa/2-1/2qa-1e-σ(a∨)) =∑y∈L∨Ay e-σ(y)$ where ${L}^{\vee }$ is the lattice in $A$ spanned by the ${a}^{\vee }$ for $a\in {\Sigma }_{0},$ and all but finitely many of the coefficients ${A}_{y}$ are zero. Then $c(s)(s,t)= ∑Ayeσ(x+p-y) ∑w∈W0ε(w)ewσ(p)$ where $x=\text{t(0)}$ and $p=\sum _{a\in {\Sigma }_{0}^{+}}{a}^{\vee },$ and $\epsilon \left(w\right)$ is the signature $\left(±1\right)$ of $w\in {W}_{0}\text{.}$ So we have $∑w∈W0c (ws)(ws,t)= ∑yAy ∑w∈W0ε(w)ewσ(x+p-y) ∑w∈W0ε(w)ewσ(p)$ and therefore the limit of the left-hand side as $s\to 1,$ that is as $\sigma \to 0,$ is $∑yAy π(x+p-y)π(p)$ where $\pi$ is the polynomial function $\prod _{a\in {\Sigma }_{0}^{+}}a$ on $A\text{.}$

To express this result in a concise form let us make the following convention: if $F=\sum {\alpha }_{i}{t}_{i}\in ℂ\left[T\right],$ where ${\alpha }_{i}\in ℂ$ and ${t}_{i}\in T,$ then $\pi \circ F$ denotes the polynomial function $\sum {\alpha }_{i}\pi \circ {t}_{i}$ on $A\text{:}$ that is, $(π∘F)(x)=∑ αiπ(ti(x)) (x∈A).$ Then we have proved

Let $t=\nu \left(z\right)\in {T}^{++}$ and put $x=t\left(0\right)\text{.}$ Let $F=∏a∈Σ0+ (1+qa/2-1/2ta-1) (1-qa/2-1/2qa-1ta-1) ∈ℂ[T].$ Then $ω1(z-1)= δ(t)-1/2Q(q-1) (π∘F)(x+p) π(p)$ where $\pi =\prod _{a\in {\Sigma }_{0}^{+}}a$ and $p=\sum _{a\in {\Sigma }_{0}^{+}}{a}^{\vee }\text{.}$

Hence ${\omega }_{1}\left({z}^{-1}\right)$ is of the form $\Phi \left(x\right){e}^{-\lambda \left(x\right)},$ where $\Phi$ is a polynomial and $\lambda$ is a linear form on $A$ such that $\lambda \left(x\right)>0$ for all $x\in {\stackrel{‾}{C}}_{0}$ (except $x=0\text{).}$ It follows that ${\omega }_{1}\left({z}^{-1}\right)\to 0$ as $x\to \infty$ in any direction in the cone ${\stackrel{‾}{C}}_{0}$ and consequently that the spherical function ${\omega }_{1}$ is bounded.

The case of arbitrary singular $s$ is quite analogous, and we shall merely state the result. Let ${\Sigma }_{0s}$ be the subsystem of ${\Sigma }_{0}$ consisting of all roots $a$ such that $|s\left({t}_{a}\right)|=1\text{;}$ put $cs(s) = ∏a∈Σ0+a∉Σ0s+ c0(a,s) Fs = ∏a∈Σ0s+ (1+qa/2-1/2ta-1) (1-qa/2-1/2qa-1ta-1) πs = ∏a∈Σ0s+ a,ps= ∑a∈Σ0s+ a∨.$

We have, for all ${s}_{0}\in S=\text{Hom}\left(T,{ℂ}^{*}\right),$ and $z\in {Z}^{++},$ $ωs0(z-1)= δ(t)-1/2 Q(q-1) ∑s (πs∘Fs) (x+ps) πs(ps) cs(s)(s,t)$ summed over all $s$ in the orbit ${W}_{0}{s}_{0}$ of ${s}_{0}\text{.}$ As before, $t=\nu \left(z\right)$ and $x=t\left(0\right)\text{.}$

Hence ${\omega }_{{s}_{0}}\left({z}^{-1}\right)$ is of the form $(4.6.3) ωs0(z-1)= ∑sΦs(x) (sδ-1/2,t)$ where the ${\Phi }_{s}$ are polynomial functions on $A,$ depending on $s\text{.}$

### Bounded spherical functions

Let ${s}_{0}\in S=\text{Hom}\left(T,{ℂ}^{*}\right)\text{.}$ As in (3.3.13) we shall identify $\text{log} |{s}_{0}|\in \text{Hom}\left(T,ℝ\right)$ with an element of ${A}^{*},$ namely the linear form on $A$ whose value at $t\left(0\right)$ is $\text{log} |{s}_{0}\left(t\right)|$ for all $t\in T\text{.}$ In particular, with this identification it makes sense to talk of the convex hull in A* of the set $D={log wδ1/2:w∈W0}.$

The spherical function ${\omega }_{{s}_{0}}$ is bounded $⇔\text{log} |{s}_{0}|$ lies in the convex hull of $D\text{.}$ Proof. Let ${C}_{0}^{*},$ ${{C}_{0}^{*}}^{\perp }$ be the images in ${A}^{*}$ of the cones ${C}_{0},$ ${C}_{0}^{\perp }$ respectively, under the isomorphism of $A$ onto ${A}^{*}$ defined by the scalar product. Then $\stackrel{‾}{{C}_{0}^{*}},$ the closure of ${C}_{0}^{*},$ is a fundamental region for the action of ${W}_{0}$ on ${A}^{*}\text{.}$ Since ${\omega }_{{s}_{0}}={\omega }_{w{s}_{0}}$ for all $w\in {W}_{0},$ we may assume that $\text{log} |{s}_{0}|\in \stackrel{‾}{{C}_{0}^{*}},$ or equivalently that $|{s}_{0}\left({t}_{a}\right)|\ge 1$ for all $a\in {\Sigma }_{0}^{+}\text{.}$ It is not hard to show that $D∩C0+‾= (log δ1/2-C0*⊥) ∩C0*‾$ so that we have to prove that $ωs0 is bounded⇔ log δ1/2- log |s0|∈ C0*⊥$ or equivalently that $ωs0 is bounded⇔ |(δ-1/2s0,t)| ≤1for all t∈T++.$ Suppose first that ${\omega }_{{s}_{0}}$ is hounded. Let $z\in {Z}^{++},$ $t=\nu \left(z\right),$ $x=t\left(0\right)$ and assume that $x\in {C}_{0},$ i.e. that $x$ does not lie on a wall of the chamber ${C}_{0}\text{.}$ By (4.6.3) we have $ωs0(z-n) = ∑s∈W0s0 Φs(nx) (δ-1/2s,tn) = (δ-1/2s0,tn) { Φs0(nx)+ ∑s≠s0 Φs(nx) (s,tn)(s0,tn) } .$ Consider the terms in this sum. We have $s=w{s}_{0}\ne {s}_{0},$ hence $|(s,tn)(s0,tn)|= |(ws0,tn)(s0,tn)| =e-nλ(x) say,$ where $\lambda =\text{log} |{s}_{0}|-w\text{log} |{s}_{0}|\in {{C}_{0}^{*}}^{\perp }\text{.}$ Since $x\in {C}_{0}$ it follows that $\lambda \left(x\right)>0\text{.}$ Also ${\Phi }_{s}\left(nx\right),$ for fixed $x,$ is a polynomial in $n\text{.}$ Hence, as $n\to \infty ,$ we have $|Φs(nx)(s,tn)(s0,tn)| =|Φs(nx)| e-nλ(x)→0$ and therefore $ωs0(z-n)∼ (δ-1/2s0,tn) Φs0(nx).$ The polynomial ${\Phi }_{{s}_{0}}$ is not identically zero. Since ${\omega }_{{s}_{0}}$ is bounded it follows that, for almost all $t\in {T}^{++},$ we have $|(δ-1/2s0,t)| ≤1.$ It follows then that $|\left({\delta }^{-1/2}{s}_{0},t\right)|\le 1$ for all $t\in {T}^{++}\text{.}$ Conversely, suppose that ${s}_{0}$ satisfies this condition. Since the set of bounded spherical functions is closed in the set $\Omega$ of all spherical functions (see e.g. Tamagawa [Tam1960], p. 378), we may assume that ${s}_{0}$ is nonsingular. But then we can use the original formula (4.1.2): $ωs0(z-1)= (δ-1/2s0,t) Q(q-1) { c(s0)+ ∑w≠1 (ws0,t) (s0,t) c(ws0) } .$ Since ${s}_{0}\in {C}_{0}^{*}$ and $t\in {T}^{++}$ we have $|(ws0,t)|≤ |(s0,t)|$ for all $w\ne 1,$ and therefore $|ωs0(z-1)| ≤1Q(q-1) ∑w∈W0 |c(ws0)|.$ Consequently ${\omega }^{{s}_{0}}$ is bounded. $\square$

## 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.