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 withâ€¨ ${w}_{a},$ and therefore $q\left(w\right)=q\left(w\prime \right)q\left({w}_{a}\right)$ by (3.1.7). Hence we haveâ€¨ $(4.5.1) q(w)Iw(s)=- q(w′)Iw′ (s).$ 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.

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