By Biasi C., de Mattos D.

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S. DeBacker 54 For example, for GL2(k) we have Let Z(G) 5 T denote the center of G. For 1 5 k 5 n define Xk E = Hom(kx,T ) by setting X,(T) (&(S))aj = { s ifi=j=k #k 1 if z = j 0 otherwise, for s E k x . With respect to our choice of a Chevalley basis, we can identify A = d ( T ) ,the apartmentb corresponding to T , with the real vector space (X*(T) @ R ) / ( X * ( Z ( G ) @ ) N. Let 20 denote the origin in A. The apartment of T is an (n- 1)-dimensional Euclidean space spanned by the set (20 xk 1 1 5 k 5 n}.

Proof: If v is in V ( N ) ,then v is a finite linear combination of terms 7r(n)u- u. Let Nj be sufficiently large so that it contains all n appearing in that linear combination. Then the integral over Nj will be 0. 0 Exercise. Prove the other direction of the Lemma. Hint: The vector v is invariant under some N i . Now assume that V is a smooth G-module. Then VN is naturally a smooth 2’-module, and V --+ VN is called the Jacquet functor. The Jacquet functor and the parabolic induction are related by the Frobenius reciprocity: H O ~(V, G I n d g (x))= H O ~( TV , ,6 1 ’ 2 ~ ) Exercise.

Proof: To prove this proposition, we need a result on the asymptotics of matrix coefficients of V . Recall that A,= ( m0a O ). 2: Let V be the Steinberg representation of G. Then, for every v in V , and V in there exists a positive integer i such that v, ("(kX,+ilC')V, 6)= 6(Xa)(7r(kXik')W,5). 2. Let U and 0 be the finite dimensional subspaces generated by v and V over K , respectively. Let u be in U , and for every positive integer a , define u, = 7r(Aa)u- b(X,)u. G. Recall that V ( N )is a union of V ( N j ) ,where V ( N j )is the kernel of the operator Pj(u) = / N j 7r(n)ud n .