A NULLSTELLENSATZ FOR AMOEBAS

ENDOSCOPIC LIFTING 455
and only if the integral
∫
θ τ (vy 4mn )ψ m (y 4mn ) dv dy 4mn
is zero for every choice of data.
Recall that V is the unipotent radical of the parabolic subgroup Q which was
defined in the beginning of the proof of Proposition 1. Arguing as in [GRS5, Theorem 1,
pages 889 – 894], we conclude that the above Fourier coefficient is nonzero for some
choice of data. Thus, we obtain a contradiction.
We also need the following local result.
PROPOSITION 2
Let F be a nonarchimedean local field. For this proposition only, let θ τ denote the
local constituent of τ at the place F . Assume that θ τ is unramified. Let l denote a
functional defined on θ τ which satisfies the property l(vy 4mn ω) = ψm −1(y
4mn)l(ω) for
all v ∈ V , for all y 4mn that were defined in the proof of Theorem 1, and for all vectors
ω ∈ θ τ . Then the space of such functionals is 1-dimensional.
Proof
Let L denote the maximal unipotent subgroup of Sp 4mn . Thus, every element in L
has a unique factorization as vy 4mn .From[G2, Lemma 3.1], we deduce that θ τ is a
quotient of the induced representation Ind Sp 4mn
̂Q (χ 1 ⊗···⊗χ m ) δ 1/2 . Here ̂Q 1
̂Q is the
standard parabolic subgroup of Sp 4mn whose Levi part is GL 2n ×···×GL 2n ,and
χ i are certain unramified characters. To prove the proposition, we apply the Bruhat
theory; thus, it is enough to prove the following. We say that g ∈ ̂Q\Sp 4mn /L is
not admissible if we can find an element y 4mn as above, so that ψ m (y 4mn ) ≠ 1 and
gy 4mn g −1 ∈ ̂Q. Otherwise, we say that g is admissible. To prove the proposition, it is
enough to show that there is at most one admissible double coset. This proves that the
space of such functionals is at most 1-dimensional. From Theorem 1, it follows that
the space of such functionals is exactly 1-dimensional.
We can choose elements in ̂Q\Sp 4mn /L as Weyl elements modulo from the left by
Weyl elements in GL 2n ×···×GL 2n . We choose the Weyl elements to be permutation
matrices. Let w be such a Weyl element. Consider the first n rows of w. We claim that
for w to be admissible, we must have w i,2ni ≠ 0 whenever 1 ≤ i ≤ n. Indeed, if not,
then from the definition of ψ m , we can find a matrix in y 4mn so that ψ m (y 4mn ) ≠ 0
and wy 4mn w −1 ∈ ̂Q. Indeed,iffor1 ≤ i ≤ n we have w i,j ≠ 0 and j ≠ 2ni, then
the matrix x(r) = I + re j,j+1 satisfies wx(r)w −1 ∈ ̂Q and ψ(x(r)) ≠ 1. Here I is
the (4mn)-identity matrix, and e p,q is the (4mn)-matrix with 1 at the (p, q)-entry and
zero elsewhere.