Degenerate Principal Series

XI

We note that the components of ny have one-dimensional spaces of Mx-fixed vectors.

By considering the central characters of the components of 7Tt/, one can show that

only two of them contain x - 1 - The question is whether they arise as a direct sum

(to get dim=2) or only as composition factors (so dim=l) . For s near the point in

question (either 0 or in/lnq)^ the action of H(M,MX) diagonalizes. It turns out

that the eigenvectors may be analytically continued, so the action of H(M,MX) on

(Vu)M* will diagonalize at the point in question. Thus, dim=2. The Hecke algebra

results are summarized in Theorem 2.7.2, which we now state. If n 2, TT = %GAX

is reducible if and only if \ is of the form | • | ± n or \ 2 — 1- If ^ = 1, in which case

the group is just Sl2(F), reducibility occurs if and only if x 1S OI" the form \ — I * \±l

or x2 = 1 with x / 1 . When reducible, TC has two components.

The computations required to use the Hecke algebra approach are too

complicated for an arbitrary maximal parabolic subgroup, so in chapter 3 we intro-

duce a different approach based on a technique of Tadic. As an example, suppose

that 7r=IndpX is a degenerate principal series representation of Sp2n{F) and that

the Jacquet module of x, treating x a s a representation of M, is a regular char-

acter on the split torus A of G. The basic idea for showing irreducibility is as

follows. Let rAG(^)ss denote the semisimplification of the Jacquet module of w

with respect to the minimal parabolic subgroup. For Pi an intermediate parabolic

subgroup, let r ^0(^)33 denote the semisimplification of the Jacquet module com-

puted with respect to Pz (see chapter 1 for a summary of the Bernstein-Zelevinsky

notation for Jacquet modules). Suppose that ipi,i/2 € rAG(ft)ss Now, suppose

that we can choose Pi so that T A ^ G ^ ^ S has an irreducible component a such

that both ipi and ip2 are in rAM^^ss- Then, if 7r0 is a composition factor of 7r

with tpi G 7\4G(7TO)S5, we claim that ip2 € rAG(^o)ss as well. To see this, look at

^MIG(7TO)S5- Since %j)X G rAG{^o)ss, n^G^o)** will have to contain at least part

of cr-by regularity, there is only one copy of ipi around. Since a is irreducible,

^MIG(TTO)55 must contain all of a, so that ^1,^2 € rAG(^o)ss We then argue as

follows: suppose xfri G rAG(^o)ss- By working with rM1G(n0)ss, we conclude that

/2 € rAG(no)ss- Then, by using rM2G(7r0)ss, we conclude that ip3 G rAG(^o)ss, etc.