A NULLSTELLENSATZ FOR AMOEBAS

ENDOSCOPIC LIFTING 471
As with the group T 0 , we suppress j from the notation. Notice that S 0 is, in fact, a
subgroup of the group of matrices of the form β(R). The function θ ɛ is left-invariant
under rational points. Thus, given a character ν as above, one can find an element
s 0 (ν) ∈ S 0 (F ) such that when we conjugate the above integral by that element from
left to right, we can collapse summation and integration in such a way that we obtain
the integral
∫
θ ɛ
(
t0 β(R)u 1 u ) ψ m,n,j (u)χ(R) dt 0 dR du 1 du
as inner integration to the above integral. Here the variable R is integrated over the
group S 0 (A)B ′ (F )\B ′ (A), where B ′ ={β(R) :R ∈ Mat j×2n }. Thus, to prove that
(10) is zero, it is enough to prove that the above integral is zero.
We proceed by induction. For 1 ≤ l ≤ m − 1, we define the following 2m − 1
families of abelian unipotent subgroups of Sp 2m(2n+1) . Here a = 2(mn + m + n):
T l =
S l =
{ 2nl+j
∑
I +
k=1
{ 2nl+j
∑
I +
k=1
}
r k e ′ 2n(l−1)+j+1,k : r 1 = r 2ns+j+1 = 0; 0 ≤ s ≤ l − 1 ,
}
p k e ′ k,2nl+j+1 : p 1 = p 2ns+j+1 = 0; 0 ≤ s ≤ l − 1 ,
{ a−j
∑
}
T m = I+ r k e ′ 2mn+m−n+1,k : r 1 = r 2ns+j+1 = r 2mn+m−n+1 = 0; 0 ≤ s ≤ l−1 ,
k=1
{ a−j
∑
}
S m = I + p k e ′ k,a−j+1 : p 1 = p 2ns+j+1 = p 2mn+m−n+1 = 0; 0 ≤ s ≤ l − 1 ,
k=1
{ a−j+2nl
∑
T m+l = I + r k e ′ a+2n(l−1)−j+1,k : r 1 = r 2ns+j+1 = 0;
k=1
}
r 2mn+m−n+1 = r a+2nq−j+1 = 0; 0 ≤ s ≤ m − 2; 0 ≤ q ≤ l − 1 ,
{ a−j+2nl
∑
S m+l = I + p k e ′ k,a+2nl−j+1 : p 1 = p 2ns+j+1 = 0;
k=1
}
p 2mn+m−n+1 = p a+2nq−j+1 = 0; 0 ≤ s ≤ m − 2; 0 ≤ q ≤ l − 1 .
All of the above groups depend on the parameter j, which we omit from the notation.