Last update: 19 February 2014
For each let be the translation defined by (so that Let and define whenever the denominator is non-zero. (If then is taken to be Likewise for If we put So if we have whilst if we have Finally put Then is defined whenever for all The function is the counterpart of Harish-Chandra's for a real semisimple Lie group.
|(3)||where is the element of which sends every positive root to a negative root.|
(1) is clear since for all
(2) is easily checked, remembering that (3.1.6).
(3) follows from (1) and the definitions.
The main result of this chapter is the following:
Let be a homomorphism such that for all Then if we have where and
The proof will occupy sections (4.2)-(4.5). To see what the formula looks like in a particular case, take as in (2.7.7). Here and for all roots where is the number of elements in the residue field of The roots are in the notation of (2.2.2) and we may write where the are non-zero complex numbers, determined up to a common factor by Then Let be a generator of the maximal ideal, then we may take to be the diagonal matrix with (and Then also by (3.2.11) where and So we have summed over all permutations of
From (3.3.1) we have where We shall prove (4.1.2) by computing this integral explicitly.
First we remark that is constant on each double coset For fixed the set meets only finitely many of these double cosets (because they are open and mutually disjoint, and is compact), and therefore the integrand in (4.2.1) takes only finitely many values, each of which is of the form for some Hence is of the form for some independent of
Next, since is constant on each double coset we may assume that Write and so that Then
Consider the integral We have to look at as runs through Now hence and therefore Hence the integration is effectively over the set or equivalently (since over where
Next, we have by (2.6.4), so that the integration is effectively over the set where is any element of and where is a translate of the cone So we have where is the Haar measure on normalized as in (3.2), and is a constant independent of To evaluate we may put so that is identically Then the left-hand side of (4.2.3) is equal to the volume of which is in the notation introduced in (3.1), and the right-hand side is equal to so that But also and therefore (4.2.3) becomes
So if we put where then from (4.2.2) and (4.2.4) we have
Since is compact it intersects only finitely many of the double cosets and so as before we conclude that
To evaluate we shall compute the integrals First of all, if then and contains the origin so that and therefore for all Consequently (because the volume of is
Now assume that Then there exists a simple root such that The group is the semi-direct product of and the group Hence is the semi-direct product of the group where and the group Hence where are the Haar measures on and respectively, normalized as in (3.2).
In order to compute we write as the difference between the sets and and consider the integrals
Here the integration is no longer over a compact set, since neither nor is compact, and we shall therefore have to restrict in order that the integrals shall converge.
Let and let be such that Then the integral converges, and its value is equal to in the notation of (4.1).
We split up into the union of and the subsets If then and therefore Consequently
Next consider From (2.6.6) we can write where and Hence because Choosing such that we have and therefore Hence and by (3.1.5) the volume of is equal to Consequently the value of the integral is making use of (3.2.7). This is a geometric series with common ratio hence converges if and only if Summing the series, we get the value announced in (4.3.1).
Consider first the integral defined in (4.2.10). We recall that in the definition of is a simple root such that
Suppose that Then the integral converges and its value is where
Choose such that Since we have Write in the form where Then where and Hence can be written uniquely in the form and therefore since normalizes and is invariant on the right with respect to So, putting we have and Hence As runs through the subgroup defined in (4.2), runs through and so we have But and by (4.3.1) this integral converges, since Since is compact it follows that converges, and we have since by (3.2.9)
To complete the proof of (4.4.1) we shall show that
Next we consider the integral where if
Suppose that Then the integral converges and is equal to where as before
As in (4.3) put so that is the disjoint union of the sets for Consider the integral The calculation proceeds on similar lines to those of (4.3). By (2.6.6) we may write with and Hence where because and
Hence But normalizes and does not change the measure, so that where as before Consequently Summing this geometric series, which converges if and only if and using (4.4.3) now leads to the result. [Recall that is an even integer if by (3.1.4).]
These calculations lead to the following reduction formula:
If then where is such that is negative, and
If this follows directly from (4.4.1) and (4.4.4). By (4.2.7) the are polynomials in and where 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
This reduction formula is the key result. In the next section we shall use it first to prove that for all which was postponed from Chapter III (3.3.3), and then to complete the proof of the formula (4.1.2).
We shall begin by proving (3.3.3). Clearly it is enough to show that for all and we may assume that otherwise and there is nothing to prove.
Put and suppose first that is a negative root, and Then by (4.4.5) where Replacing by and subtracting, we get by (4.1.1). Since is negative, there is a reduced word for ending with and therefore by (3.1.7). Hence we have If on the other hand is positive, then the roles of and are interchanged, and (4.5.1) is therefore true for all Summing over it follows that and hence by (4.2.6).
Let and let be a reduced word in the generators Let be the set of all products where is or
If for all we shall say that is nonsingular.
Suppose is nonsingular. Then with coefficients independent of Also the product being taken over the roots such that is negative.
by induction on the length of If then and by (4.2.8). If apply (4.4.5) with
From (4.5.2) and (4.2.6) it follows that we have with coefficients independent of Consider in particular the coefficient where is the unique element of which sends every positive root to a negative root. This coefficient comes only from and hence by (4.5.2) we have since
Next we have
Assume that is nonsingular (i.e. that for all Then the homomorphisms with are linearly independent over
Assuming this for the moment, since for all we have from (4.5.3) and therefore (assuming nonsingular) by (4.5.5) for all In particular, therefore by (4.5.4); or, replacing by by (4.1.1). Hence (4.5.3) now becomes provided that is nonsingular. However, this restriction is superfluous, for the left-hand side of (4.5.6) is of the form for some and the right-hand side is of the form for some the field of fractions of Since and take the same values on a dense open set in it follows that they are equal and therefore (4.5.6) is valid whenever the right-hand side is defined, i.e. provided is nonsingular.
Finally, to derive the formula (4.3.2) we have merely to replace by and put in (4.5.6), bearing in mind (3.3.2).
There remains the proof of (4.5.5). This will follow from
Let be distinct elements of Then they are linearly independent.
The argument is a familiar one. If the are linearly dependent, choose a linear relation between them containing as few non-zero terms as possible, say (renumbering the if necessary) with each Since there exists such that Then we have and for all so by subtraction we get contradicting the minimality of
From (4.5.7) it follows that if the set is linearly dependent, then for some in and hence is fixed by an element in Hence the same is true of Identifying with an element as in (3.3.13), it follows that has non-trivial stabilizer under the action of on and hence lies on a reflecting hyperplane. In other words, we have for some and this means that This proves (4.5.5) and hence completes the proof of (4.1.2).
If (which is nonsingular) then is identically Hence also by (3.3.2). Putting in (4.1.2) gives therefore for all Consider for There exists a simple root such that is negative. Let then we have and therefore so that and therefore Consequently the sum above reduces to the term corresponding to that is This suggests (but does not prove) the following multiplicative formula for the Poincare polynomial of the root system For define Then with the convention that if In fact (4.5.9) is true and can be proved by elementary methods. Putting for all and assuming that is reduced (i.e. that (4.5.9) specializes to give (since and have the same Weyl group
Next we consider the problem of finding an expression for when is singular, i.e. when for some We have remarked earlier that is of the form (where for some polynomial independent of Hence we can calculate for singular as the limit of as through nonsingular values of 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 Then we have to evaluate the limit as of In calculating the limit, we may make through real values. So let be real and let be the linear form on defined by for all Then so that Let us temporarily write where is the lattice in spanned by the for and all but finitely many of the coefficients are zero. Then where and and is the signature of So we have and therefore the limit of the left-hand side as that is as is where is the polynomial function on
To express this result in a concise form let us make the following convention: if where and then denotes the polynomial function on that is, Then we have proved
Let and put Let Then where and
Hence is of the form where is a polynomial and is a linear form on such that for all (except It follows that as in any direction in the cone and consequently that the spherical function is bounded.
The case of arbitrary singular is quite analogous, and we shall merely state the result. Let be the subsystem of consisting of all roots such that put
We have, for all and summed over all in the orbit of As before, and
Hence is of the form where the are polynomial functions on depending on
As in (3.3.13) we shall identify
with an element of namely the linear form on whose value at
for all In particular, with this identification it makes sense to talk of the convex hull in
The spherical function is bounded lies in the convex hull of
Let be the images in of the cones respectively, under the isomorphism of onto defined by the scalar product. Then the closure of is a fundamental region for the action of on
Since for all we may assume that or equivalently that for all It is not hard to show that so that we have to prove that or equivalently that
Suppose first that is hounded. Let and assume that i.e. that does not lie on a wall of the chamber By (4.6.3) we have Consider the terms in this sum. We have hence where Since it follows that Also for fixed is a polynomial in Hence, as we have and therefore The polynomial is not identically zero. Since is bounded it follows that, for almost all we have It follows then that for all
Conversely, suppose that satisfies this condition. Since the set of bounded spherical functions is closed in the set of all spherical functions (see e.g. Tamagawa [Tam1960], p. 378), we may assume that is nonsingular. But then we can use the original formula (4.1.2): Since and we have for all and therefore Consequently is bounded.
This is a typed version of the book Spherical Functions on a Group of 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.