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Моу , Representations of U{2,1) over a p-adic field

Observe that these elements are conjugate to (3. 4). In all cases the extension of Q^ to KJKi^^ is represented by a nilpotent element n. This element n has rank 2 in cases (3. 1 a, b) and rank 1 in cases (3. 1 c, d, e).

Let и and U be, as in section 2, the set of upper and lower triangular unipotent elements in В respectively. A straightforward calculation shows that under conjugation the sets (3. la—e) are fixed by U and the sets (3. la, c, d) are fixed by U, To determine how the sets (3. 1 b, e) behave under conjugation by U, write an element w of [/ as

■ 1 0 0-

nX 10 ,nY -nX 1.

XX

X in R, and Y=vy8 —n v in R

If ä is the set (3. lb), then uäu Ms the set (3.1b*)

- i - l

0 1 0

K { a - bv ] / s ) 0 -1

0 -n{a-v\/^) 0_

If ä is the set (3. le), then иаи~^ is the set

( 3 . le*)

n { a\ / ^ + sv) 0 |/e

0 7ca|/ë 0

P P R

p2 p p p2 p2 p

- i - l

+

P P R'

p2 p p p2 p2 p

A criterion for determining the support of the Hecke algebra Ж{С//Ь, QJ is given by the following analog of Lemma 2. 1. Its proof is exactly as in Lemma 2. 1 and is thus omitted.

Lemma 3. L An element g in G lies in the support of Ж{С//h, Q^ if ana only if the intersection Adg.a n a is nonempty.

We now use the lemma to show the support of Ж{С//Ь, QJ is contained in a compact group.

Theorems . 2. The support of Ж{С//Ь, ÜJ is contained in K' in cases (3. lb), (3. le) and contained in В in cases (3. la), (3. Ic), (3. Id).

Proof . Consider first cases (3. 1 a), (3. 1 c) and (3. 1 d). Suppose g in supp Jf (G//L, ß„). By the sharper form (2. 4 a, b) of Bruhat's decomposition explained in section 2, write g as wdwu', where и and u' belong either to (7 or I/ according to the type of w. Since Adw and Adu' fix the set a, the intersection Adg.a n a is nonempty if and only if (dw)~^ a(d>v) n a is nonempty. If the element tfw is a diagonal element and not in A{R\ then Ad(dw) will stretch the strictly upper and shrink the strictly lower entries or vice versa. In both cases Ad(dw).a does not intersect a. If dw is antidiagonal, then Ad(dw) will exchange the entry in the first row third column with the entry in the third row first column and change the valuation of the entries by an even integer. This means Ad(dw).a does not intersect a. We conclude that g must lie in B. The proof of the theorem in cases (3. lb) and (3. le) is essentially the same but more complicated