Spherical Functions on a Group of p-adic Type

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 19 February 2014

Chapter IV: Calculation of the spherical functions

Statement of result

For each bΣ012Σ0 let tbT be the translation defined by tb(0)=b (so that ta/2=ta2). Let sHom(T,*) 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 bΣ1, then qb is taken to be 1. Likewise for qb/2.) If aΣ0 we put c0(a,s)= c(12a,s) c(a,s). So if 12aΣ1 we have c0(a,s)= c(a,s)= 1-qa-1 s(ta)-1 1-s(ta)-1 whilst if 12aΣ1 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)= aΣ1+ c(b,s)= aΣ0+ c0(a,s). Then c(s) is defined whenever s(tb)1 for all bΣ1. The function c is the counterpart of Harish-Chandra's c-function for a real semisimple Lie group.

(1) c(b,s)=c(wb,ws) for bΣ1 and wW0.
(2) c0(a,s)+c0(-a,s)=1+q(wa)-1.
(3) c(s-1)=c(w0s), where w0 is the element of W0 which sends every positive root to a negative root.


(1) is clear since qwb=qb for all wW0.

(2) is easily checked, remembering that q(wa)=qa/2qa (3.1.6).

(3) follows from (1) and the definitions.

The main result of this chapter is the following:

Let s:T* be a homomorphism such that s(tb)1 for all bΣ1. Then if z0Z++ we have ωs(z0-1)= δ(t0)-1/2 Q(q-1) wW0c (ws)(ws,t0) where t0=ν(z0) and Q(q-1)=wW0q(w)-1.

The proof will occupy sections (4.2)-(4.5). To see what the formula looks like in a particular case, take G=SLl+1(𝓀) as in (2.7.7). Here Σ1=Σ0 and qa=q for all roots a, where q is the number of elements in the residue field of 𝓀. The roots are ei-ej (ij) in the notation of (2.2.2) and we may write s(tei-ej)=ξiξj-1, where the ξi are non-zero complex numbers, determined up to a common factor by s. Then c(s)=i<j 1-q-1 ξi-1 ξj 1-ξi-1ξj =i<j ξi-q-1ξj ξi-ξj . Let π be a generator of the maximal ideal, then we may take z0 to be the diagonal matrix z0= ( πλ0 πλl ) with λ0λ1λl (and λi=0). Then (s,t0)=ξ0λ0ξlλl; also by (3.2.11) δ(t)1/2=qn(λ) where n(λ)=lλ0+(l-1)λ1++λl-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<j ξσ(i)-q-1ξσ(j) ξσ(i)-ξσ(j) summed over all permutations σ of {0,1,,l}.

Preliminary reductions

From (3.3.1) we have (4.2.1) ωs(g)= Kϕs (g-1k) dk where ϕs(kzu-)=(sδ1/2,ν(z)). We shall prove (4.1.2) by computing this integral explicitly.

First we remark that ϕs is constant on each double coset KtU-. For fixed gG, the set g-1K meets only finitely many of these double cosets (because they are open and mutually disjoint, and g-1K is compact), and therefore the integrand in (4.2.1) takes only finitely many values, each of which is of the form (sδ1/2,t) for some tT. Hence ωs(g) is of the form s(Φg) for some Φg[T] independent of s.

Next, since ωs is constant on each double coset KtK we may assume that gZ++. Write g=z-1 and t=ν(z), so that t-1T++. Then (4.2.2) ωs(z-1) = Kϕs (zk)dk = wW0 BwB ϕs(zk) dk.

Consider the integral BwBϕs(zk)dk. We have to look at ϕs(g) as g runs through zBwB=zU(C0)wB=Ut(C0)twB. Now t-1(0)C0, hence 0t(C0) and therefore Ut(C0)K. Hence the integration is effectively over the set twB, or equivalently (since wK) over tB where t=w-1tw.

Next, we have B=U(C0)HU(C)- by (2.6.4), so that the integration is effectively over the set zU(C0), where z is any element of ν-1(t); and zU(C0)=U(C0)z where C0=t(C0) is a translate of the cone C0. So we have (4.2.3) BwBϕs (zk)dk=κ U(C0) ϕs(u+z) du+ where du+ is the Haar measure on U+, normalized as in (3.2), and κ is a constant independent of s. To evaluate κ we may put s=δ-1/2, so that ϕs is identically 1. Then the left-hand side of (4.2.3) is equal to the volume of BwB, which is q(w)/Q(q) 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 ϕs(u+z)=ϕs(u+)·(sδ1/2,t), 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 C0=t(C0)=w-1tw(C0), then from (4.2.2) and (4.2.4) we have (4.2.6) ωs(z-1)= 1Q(q) wW0q(w) Jw(s).

Since U(C0) is compact it intersects only finitely many of the double cosets KtU-, and so as before we conclude that (4.2.7) Jw(s)=s (Φz,w) for someΦz,w (T).

To evaluate ωs(z-1) we shall compute the integrals Jw(s). First of all, if w=1, then t=t and C0=t(C0) contains the origin 0, so that U(C0)K and therefore ϕs(u+)=1 for all u+U(C0). Consequently (4.2.8) J1(s)= (sδ1/2,t) (because the volume of U(C0) is δ(t)).

Now assume that w1. Then there exists a simple root aΠ0 such that waΣ0-. The group U(C0) is the semi-direct product of Ua and the group Ua=Ub:bΣ0+,ba. Hence U(C0)=zU(C0)z-1 is the semi-direct product of the group V=zUaz-1 U(a) where U(a)= U(b):bΣ0+,ba, and the group zUaz-1= Ut(a). Hence (4.2.9) Jw(s)= (w(sδ-1/2),t) VUt(a) ϕs(uaua)d uadua where dua, dua are the Haar measures on U(a) and U(a) respectively, normalized as in (3.2).

In order to compute Jw(s) we write Ut(a) as the difference between the sets U(a) and U(a)-Ut(a), and consider the integrals (4.2.10) { Jw1(s)= (w(sδ-1/2),t) VU(a) ϕs(uaua) duadua, Jw0(s)= (w(sδ-1/2),t) VU(a)-Ut(a) ϕs(uaua) duadua.

Here the integration is no longer over a compact set, since neither U(a) nor U(a)-Ut(a) is compact, and we shall therefore have to restrict s in order that the integrals shall converge.

The integral U(a)ϕs(ua)dua

Let aΣ0+ and let s:T* be such that |s(ta)|>1. Then the integral U(a) ϕs(ua) dua converges, and its value is equal to c0(a,s)=c(12a,s)c(a,s) in the notation of (4.1).


We split up U(a) into the union of Ua and the subsets Yr=Ua-r- Ua-r+1 (r1). If uaUa, then uaK and therefore ϕs(ua)=1. Consequently Uaϕa (ua)dua=1.

Next consider uaYr (r1). From (2.6.6) we can write ua=u1nu2 where ua,u2U-a+r and ν(n)=wa-r. Hence ϕs(ua)= ϕs(u1nu2) =ϕs(n) because U-a+rU-K. Choosing naN such that ν(na)=wa we have naK and therefore ϕs(ua)= ϕs(nan)= (sδ1/2,wawa-r) =(sδ1/2,ta-r). Hence Yrϕs( ua)dua= (sδ1/2,ta-r) ·volume ofYr and by (3.1.5) the volume of Yr is equal to qa/2[r/2] qar- qa/2[(r-1)/2] qar-1. Consequently the value of the integral is 1+r=1s (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(ta)-2, hence converges if and only if |s(ta)|>1. Summing the series, we get the value announced in (4.3.1).

Reduction of Jw(s)

Consider first the integral Jw1(s) defined in (4.2.10). We recall that in the definition of Jw(s), a is a simple root such that waΣ0-.

Suppose that |s(ta)|>1. Then the integral Jw1(s) converges and its value is c0(a,s) Jw(was) where w=wwa.


Choose naN such that ν(na)=wa. Since naK we have ϕs(uaua)=ϕs(na-1uana·na-1ua). Write na-1uana in the form kz1u-, where kK, z1Z, u-U-. Then ϕs(uaua) = ϕs ( z1u-· na-1ua ) = ϕs ( z2·na u-na-1 ·ua ) where z2=naz1na-1, and nau-na-1naU-na-1U(-a)·U(a). Hence nau-na-1 can be written uniquely in the form u-aua (u-aU(-a), uaU(a)) and therefore ϕs(uaua)= ϕs(z2u-auaua)= ϕs(z2uaua) since U(a) normalizes U(-a) and ϕs is invariant on the right with respect to U-U(-a). So, putting ua=uaua, we have ϕs(uaua)= ϕs(z2ua)= ϕs(z2uaz2-1) (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(z2uaz2-1)· ϕwas (na-1uana)· (δ1/2,ν(z1-1z2)). As ua runs through the subgroup V=zUaz-1 defined in (4.2), na-1uana runs through V=na-1Vna and so we have Jw1(s)= (w(sδ-1/2),t) Vϕwas (ua) (δ1/2,ν(z1-1z2)) dua×U(a) ϕs(z2uaz2-1) dua. But U(a)ϕs (z2uaz2-1) 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(ta)|>1. Since V is compact it follows that Jw1(s) converges, and we have (4.4.2) Jw1(s)= (w(sδ-1/2),t) c0(a,s) Vϕwas (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 sHom(T,*), Jw(s)= (ws,t) (wδ-1/2,t) Vϕs (ua)dua.


From the definition (4.2.5), Jw(s)= (w(sδ-1/2),t) U(C0) ϕs(ua)dua where C0=(w-1tw)(C0), so that U(C0) is the semi-direct product of V and Uβ, where β=(w-1tw)a=a-wa(x) where x=t(0).

Since w=wwa, we have β(0)=a(0)+ wa(x)=wa(x) which is 0 because wa is a negative root, and x-C0. Hence UK, so that Uβϕs(ua) dua = volume ofUβ = δa(w-1tw) = δ(w-1tw)1/2/ δ(w-1tw)1/2 by (3.2.9) = (wδ1/2,t)· (wδ-1/2,t). This completes the proof of (4.4.3) and hence also of (4.4.1).

Next we consider the integral Jw0(s)= (w(sδ-1/2),t) VU(a)-Uα ϕs(uaua)d uadua where α=(w-1tw)(a)=a-wa(x) if x=t(0).

Suppose that |s(ta)|>1. Then the integral Jw0(s) converges and is equal to (c0(a,s)-1) Jw(s) where as before w=wwa.


As in (4.3) put Yr=Ua-r-Ua-r+1 so that U(a)-Uα is the disjoint union of the sets Yr for r1+wa(x)>0. Consider the integral VYrϕs (uaua)dua dua. The calculation proceeds on similar lines to those of (4.3). By (2.6.6) we may write ua=u1nu2 with u1,u2U-a+r and ν(n)=wa-r. Hence ϕs(uaua) = ϕs(uau1n) (becauseu2U-) = ϕs(u1-1uau1na-1·nan) (becauseu1K) where naν-1(wa), =ϕs (u1-1uau1na-1) ·(sδ1/2,ta-r) because nan and ν(nan)=wawa-r=ta-r.

Hence VYrϕs (uaua)dua dua= ( Vϕs (u1-1uau1na-1) dua ) ×(sδ1/2,ta-r) vol.Yr. But u1 normalizes V and does not change the measure, so that Vϕs (u1-1uau1na-1) dua = Vϕs (uana-1) dua = Vϕs (ua)dua where as before V=na-1Vna. Consequently VU(a)-Uα ϕs(uaua)d uadua= (Vϕs(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(ta)|>1, and using (4.4.3) now leads to the result. [Recall that wa(x) is an even integer if qa/21, by (3.1.4).]

These calculations lead to the following reduction formula:

If s(ta)1, then Jw(s)=c0 (a,s)Jw (was)- (c0(a,s)-1) Jw(s), where aΠ0 is such that wa is negative, and w=wwa.


If |s(ta)|>1, this follows directly from (4.4.1) and (4.4.4). By (4.2.7) the J's are polynomials in ξi=s(tai) and ξi-1 (1il), where a1,,al 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(ta)1.

This reduction formula is the key result. In the next section we shall use it first to prove that ωs=ωws for all wW0, 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 ωs(z-1)=ωwas(z-1) for all aΠ0, and we may assume that s(ta)1, otherwise s=was and there is nothing to prove.

Put Iw(s)=Jw(s)-Jw(was) and suppose first that wa is a negative root, and aΠ0. Then by (4.4.5) Jw(s)=-c0 (a,s)Iw (s)+Jw(s) where w=wwa. Replacing s by was 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
 wa, and therefore q(w)=q(w)q(wa) 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 and w are interchanged, and (4.5.1) is therefore true for all wW0. Summing over wW0, it follows that wW0q(w) Iw(s)=0 and hence ωs(z-1)=ωwas(z-1) by (4.2.6).

Let wW0 and let w=w1wr be a reduced word in the generators wa, aΠ0. Let E be the set of all products w1ε1wrεr where εi is 0 or 1.

If s(tb)1 for all bΣ1, we shall say that s is nonsingular.

Suppose s is nonsingular. Then Jw(s)=δ (t)1/2 wE λw,w(s) ·(ws,t) with coefficients λw,w(s) independent of t. Also λw,w(s)= ac0 (a,s) the product being taken over the roots aΣ0+ such that wa is negative.


by induction on the length of l(w). If l(w)=0 then w=1 and J1(s)=δ(t)1/2(s,t) by (4.2.8). If l(w)>0, apply (4.4.5) with w=w1wr-1.

From (4.5.2) and (4.2.6) it follows that we have (4.5.3) ωs(z-1)= δ(t)1/2 wW0 μw(s) (ws,t) with coefficients μw(s) independent of t=ν(z). Consider in particular the coefficient μw0(s), where w0 is the unique element of W0 which sends every positive root to a negative root. This coefficient comes only from Jw0(s) 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(w0)-1Q(q)=Q(q-1).

Next we have

Assume that |s| is nonsingular (i.e. that |s(ta)|1 for all aΣ0). Then the homomorphisms ws with wW0 are linearly independent over .

Assuming this for the moment, since ωs=ωws for all wW0 we have from (4.5.3) wW0μw (s)(ws,t)= wW0μw (ws)(wws,t) and therefore (assuming |s| nonsingular) by (4.5.5) μw(s)=μ1(ws) for all wW0. In particular, therefore μ1(w0s)= c(s)Q(q-1) by (4.5.4); or, replacing s by w0s, 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) wW0c (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(Φt) for some Φt[T], and the right-hand side is of the form s(Ψt) for some Ψt(T), the field of fractions of [T]. Since Φt and Ψt take the same values on a dense open set in S=Hom(T,*), 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 z0=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 s1,,sn (n1) be distinct elements of Hom(T,*). Then they are linearly independent.


The argument is a familiar one. If the si are linearly dependent, choose a linear relation between them containing as few non-zero terms as possible, say i=1rλi si=0 (renumbering the si if necessary) with each λi0. Since s1sr there exists t0T such that s1(t0)sr(t0). Then we have i=1rλi si(t)=0 and i=1rλisi (t0)si(t)=0 for all tT, so by subtraction we get i=1rλi (si(t0)-sr(t0)) si=0 contradicting the minimality of r.

From (4.5.7) it follows that if the set {ws:wW0} is linearly dependent, then w1s=w2s for some w1w2 in W0, and hence s is fixed by an element 1 in W0. Hence the same is true of log|s|Hom(T,). Identifying log|s| with an element uA* as in (3.3.13), it follows that u has non-trivial stabilizer under the action of W0 on A* and hence lies on a reflecting hyperplane. In other words, we have u(a)=0 for some aΣ0, and this means that |s(ta)|=1. This proves (4.5.5) and hence completes the proof of (4.1.2).

If s=δ-1/2 (which is nonsingular) then ωs is identically 1. Hence also ωδ1/2=1 by (3.3.2). Putting s=δ1/2 in (4.1.2) gives therefore Q(q-1)=δ (t0)-1/2 wW0c (wδ1/2) (wδ1/2,t0) for all t0T++. Consider c(wδ1/2) for w1. There exists a simple root a such that w-1a is negative. Let b=-w-1a, then we have tb-1=wtaw-1 and therefore wδ1/2 (tb)-1= δ1/2(ta)= qa/21/2 qa=qb/21/2 qb so that c(b,wδ1/2)=0 and therefore c(wδ1/2)=0. Consequently the sum above reduces to the term corresponding to w=1, that is Q(q-1) = c(δ12) = bΣ1+ 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(ξ) of the root system Σ1. For bΣ1, define ηb=cΣ1+ ξcc(b). Then (4.5.9) Q(ξ)= bΣ1+ 1-ξb/21/2 ξbηb1/2 1-ξb/21/2 ηb1/2 with the convention that ξb/2=1 if b2Σ1. In fact (4.5.9) is true and can be proved by elementary methods. Putting ξb=ξ for all bΣ1, and assuming that Σ1 is reduced (i.e. that Σ1=Σ0), (4.5.9) specializes to give wW0 ξl(w) = aΣ0+ 1-ξ1+ht(a) 1-ξht(a) = aΣ0+ 1-ξ1+ht(a) 1-ξht(a) (since Σ0 and Σ0 have the same Weyl group W0).

The singular case

Next we consider the problem of finding an expression for ωs(z-1) when s is singular, i.e. when s(tb)=1 for some bΣ1. We have remarked earlier that ωs(z-1) is of the form s(Ψt) (where t=ν(z)) for some polynomial Ψt[T]W0, independent of s. Hence we can calculate ωs(z-1) for singular s as the limit of ωs1(z-1) as s1s through nonsingular values of s1. 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. Then we have to evaluate the limit as s1 of wW0c (ws)(ws,t) (tT++). In calculating the limit, we may make s1 through real values. So let s be real and let σ be the linear form on A defined by σ(t(0))=logs(t) for all tT. Then s(ta)=eσ(a), so that c(s)=aΣ0+ (1+qa/2-1/2e-σ(a)) (1-qa/2-1/2qa-1e-σ(a)) 1-e-σ(2a) . Let us temporarily write aΣ0+ (1+qa/2-1/2e-σ(a)) (1-qa/2-1/2qa-1e-σ(a)) =yLAy e-σ(y) where L is the lattice in A spanned by the a for aΣ0, and all but finitely many of the coefficients Ay are zero. Then c(s)(s,t)= Ayeσ(x+p-y) wW0ε(w)ewσ(p) where x=t(0) and p=aΣ0+a, and ε(w) is the signature (±1) of wW0. So we have wW0c (ws)(ws,t)= yAy wW0ε(w)ewσ(x+p-y) wW0ε(w)ewσ(p) and therefore the limit of the left-hand side as s1, that is as σ0, is yAy π(x+p-y)π(p) where π is the polynomial function aΣ0+a on A.

To express this result in a concise form let us make the following convention: if F=αiti[T], where αi and tiT, then πF denotes the polynomial function αiπti on A: that is, (πF)(x)= αiπ(ti(x)) (xA). Then we have proved

Let t=ν(z)T++ and put x=t(0). Let F=aΣ0+ (1+qa/2-1/2ta-1) (1-qa/2-1/2qa-1ta-1) [T]. Then ω1(z-1)= δ(t)-1/2Q(q-1) (πF)(x+p) π(p) where π=aΣ0+a and p=aΣ0+a.

Hence ω1(z-1) is of the form Φ(x)e-λ(x), where Φ is a polynomial and λ is a linear form on A such that λ(x)>0 for all xC0 (except x=0). It follows that ω1(z-1)0 as x in any direction in the cone C0 and consequently that the spherical function ω1 is bounded.

The case of arbitrary singular s is quite analogous, and we shall merely state the result. Let Σ0s be the subsystem of Σ0 consisting of all roots a such that |s(ta)|=1; put cs(s) = aΣ0+aΣ0s+ c0(a,s) Fs = aΣ0s+ (1+qa/2-1/2ta-1) (1-qa/2-1/2qa-1ta-1) πs = aΣ0s+ a,ps= aΣ0s+ a.

We have, for all s0S=Hom(T,*), and zZ++, ωs0(z-1)= δ(t)-1/2 Q(q-1) s (πsFs) (x+ps) πs(ps) cs(s)(s,t) summed over all s in the orbit W0s0 of s0. As before, t=ν(z) and x=t(0).

Hence ωs0(z-1) is of the form (4.6.3) ωs0(z-1)= sΦs(x) (sδ-1/2,t) where the Φs are polynomial functions on A, depending on s.

Bounded spherical functions

Let s0S=Hom(T,*). As in (3.3.13) we shall identify log|s0|Hom(T,) with an element of A*, namely the linear form on A whose value at t(0) is log|s0(t)| for all tT. In particular, with this identification it makes sense to talk of the convex hull in A* of the set D={logwδ1/2:wW0}.

The spherical function ωs0 is bounded log|s0| lies in the convex hull of D.


Let C0*, C0* be the images in A* of the cones C0, C0 respectively, under the isomorphism of A onto A* defined by the scalar product. Then C0*, the closure of C0*, is a fundamental region for the action of W0 on A*.

Since ωs0=ωws0 for all wW0, we may assume that log|s0|C0*, or equivalently that |s0(ta)|1 for all aΣ0+. It is not hard to show that DC0+= (logδ1/2-C0*) C0* so that we have to prove that ωs0is bounded logδ1/2- log|s0| C0* or equivalently that ωs0is bounded |(δ-1/2s0,t)| 1for alltT++.

Suppose first that ωs0 is hounded. Let zZ++, t=ν(z), x=t(0) and assume that xC0, i.e. that x does not lie on a wall of the chamber C0. By (4.6.3) we have ωs0(z-n) = sW0s0 Φs(nx) (δ-1/2s,tn) = (δ-1/2s0,tn) { Φs0(nx)+ ss0 Φs(nx) (s,tn)(s0,tn) } . Consider the terms in this sum. We have s=ws0s0, hence |(s,tn)(s0,tn)|= |(ws0,tn)(s0,tn)| =e-nλ(x)say, where λ=log|s0|-wlog|s0|C0*. Since xC0 it follows that λ(x)>0. Also Φs(nx), for fixed x, is a polynomial in n. Hence, as n, 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 Φs0 is not identically zero. Since ωs0 is bounded it follows that, for almost all tT++, we have |(δ-1/2s0,t)| 1. It follows then that |(δ-1/2s0,t)|1 for all tT++.

Conversely, suppose that s0 satisfies this condition. Since the set of bounded spherical functions is closed in the set Ω of all spherical functions (see e.g. Tamagawa [Tam1960], p. 378), we may assume that s0 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)+ w1 (ws0,t) (s0,t) c(ws0) } . Since s0C0* and tT++ we have |(ws0,t)| |(s0,t)| for all w1, and therefore |ωs0(z-1)| 1Q(q-1) wW0 |c(ws0)|. Consequently ωs0 is bounded.

Notes and references

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

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

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