TRANSACTIONS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 351, Number 2, February 1999, Pages 781–811
S 0002-9947(99)02126-1

THE NONARCHIMEDEAN THETA CORRESPONDENCEFOR GSp(2) AND GO(4)BROOKS ROBERTS

Abstract. In this paper we consider the theta correspondence between thesets Irr(GSp(2, k)) and Irr(GO(X)) when k is a nonarchimedean local field anddimk X = 4. Our main theorem determines all the elements of Irr(GO(X))
that occur in the correspondence. The answer involves distinguished representations. As a corollary, we characterize all the elements of Irr(O(X)) that
occur in the theta correspondence between Irr(Sp(2, k)) and Irr(O(X)). We
also apply our main result to prove a case of a new conjecture of S.S. Kudla
concerning the first occurrence of a representation in the theta correspondence.

Suppose k is a nonarchimedean local field of characteristic zero and odd residual characteristic, X is an even dimensional nondegenerate symmetric bilinear
space over k and n is a nonnegative integer. Let ω be the Weil representationof Sp(n, k) × O(X) corresponding to a fixed choice of nontrivial additive characterof k, and let RX (Sp(n, k)) be the set of elements of Irr(Sp(n, k)) that are nonzero
quotients of ω; similarly define Rn (O(X)). By [W], the condition that π ⊗C σ be anonzero quotient of ω for π in RX (Sp(n, k)) and σ in Rn (O(X)) defines a bijection
between RX (Sp(n, k)) and Rn (O(X)). By [R], the extension of ω to the subgroup
R of GSp(n, k) × GO(X) consisting of pairs whose entries have the same similitude factor also defines a well behaved correspondence between Irr(GSp(n, k)+ )and Irr(GO(X)). Here, GSp(n, k)+ is the subgroup of elements of GSp(n, k) whose
similitude factors lie in the group of similitude factors of the elements of GO(X);
thus, GSp(n, k)+ is of index at most two in GSp(n, k) and contains Sp(n, k). To
be more precise about the correspondence, let RX (GSp(n, k)+ ) be the set of elements of Irr(GSp(n, k)+ ) whose restrictions to Sp(n, k) are multiplicity free andhave a constituent in RX (Sp(n, k)); similarly define Rn (GO(X)). Then by [R] thecondition
HomR (ω, π ⊗C σ) 6= 0
defines a bijection between RX (GSp(n, k)+ ) and Rn (GO(X)). Granted the theta dichotomy conjecture, if dimk X ≤ 2n, then GSp(n, k)+ may be replaced by GSp(n, k)
in this result.
Received by the editors June 3, 1996 and, in revised form, February 6, 1997.
1991 Mathematics Subject Classification. Primary 11F27; Secondary 22E50.
Key words and phrases. Nonarchimedean theta correspondence, GSp(2), GO(4).
During the period of this work the author was a Research Associate with the NSF 1992–
1994 special project Theta Functions, Dual Pairs, and Automorphic Forms at the Universityof Maryland, College Park, and was supported by a Stipendium at the Max-Planck-Institut fürMathematik.
c 1999 American Mathematical Society

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Since S.S. Kudla proved the compatibility of the theta correspondence with induction in [K], the characterization of RX (Sp(n, k)) and Rn (O(X)), and analogously, RX (GSp(n, k)+ ) and Rn (GO(X)), remains as a fundamental open problem. One solution might involve epsilon factors. See [HKS] for some progress inthis direction in the unitary group case. In light of the many examples relating
distinguished representations and functorality, one might also investigate whether
distinguished representations could be involved in a solution to this problem. See[GRS] for a global example, again in the case of unitary groups. Recently, S.S.
Kudla has made some new conjectures concerning the first occurrence of a representation in the theta correspondence which also shed some light on this problem.
In this paper we consider R2 (GO(X)) when dimk X = 4. Our main result is
a complete characterization of R2 (GO(X)) in this case using distinguished representations. As a corollary, we determine R2 (O(X)). Also, as a consequence of themain theorem and some other results, we prove a case of one of the above mentioned
conjectures of S.S. Kudla. Even though we obtain corollaries for the theta correspondence for isometries, we emphasize that the statements of our results and our
methods of proof depend strongly on the use of similitudes. It would be interesting
to see to what degree our results and methods extend to other situations. The employment of distinguished representations may possibly generalize. See section 4.
As far as we know, the case of Kudla’s conjecture proven in this paper is the highest
dimensional case known. We believe that one of the more valuable aspects of thiswork is to provide a model for global considerations. It would be very interesting
to determine if analogous results hold globally. A summary of previous work on
this example appears near the end of this introduction.
To state the main theorem we need some more terminology. Assume dimk X = 4.
Let π be contained in Irr(GSO(X)). If π induces irreducibly to GO(X) we say that
π is regular; otherwise, we say that π is invariant. If π is invariant, then π has two
extensions to GO(X). If y in X is anisotropic, then the stabilizer in O(X) of y canbe identified with O(Y ), where Y is the orthogonal complement to y. We say that
π is distinguished if π is invariant and there is a y such that
HomSO(Y ) (π, 1) 6= 0,
and, if disc(X) 6= 1, then Y is isotropic. We show in Corollary 7.5 that the assumption that Y is isotropic if disc(X) 6= 1 is unnecessary; however, the more restrictive
definition is convenient. Suppose that π is distinguished. Then
dimC HomSO(Y ) (π, 1) = 1.
It follows that for exactly one extension π 0 of π to GO(X) we have HomO(Y ) (π 0 , 1) 6=
0. Call this extension π + ; the other extension of π to GO(X) will be called π − . For
technical reasons, if disc(X) 6= 1, π is one dimensional and invariant, then we also
will say that π is distinguished. In this case there are also definitions of π + andπ − . The definitions of π + and π − do not depend on the choice of y. See section 4.
Theorem 6.8 (Main Theorem). Assume dimk X = 4. Let σ be in Irr(GO(X)).
Then σ is in R2 (GO(X)) if and only if σ is not of the form π − for some distinguished π in Irr(GSO(X)).
This result is entirely analogous to the case dimk X = 2n = 2 considered byHecke, Weil, Jacquet, Langlands and others. In this case, the role of SO(Y ) is
played by SO(X). For a description of this case, see section 7. Since the elements

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)783

of Irr(GO(X)) have multiplicity free restrictions to O(X) we immediately obtainthe following corollary.
Corollary 6.9. Assume dimk X = 4. Let σ1 be in Irr(O(X)). Then σ1 is in
R2 (O(X)) if and only if σ1 is not an irreducible constituent of π − |O(X) for somedistinguished π in Irr(GSO(X)).
To describe the proof and make the theorem concrete, we characterize GSO(X)in terms of units of quaternion algebras. If disc(X) = 1, there is an isomorphismof GSO(X) with either (Gl(2, k) × Gl(2, k))/k × or (D× × D× )/k × , depending on
the Hasse invariant of X. In the first case X is isotropic; in the second case, X
is anisotropic. Here, D is the division quaternion algebra over k. If disc(X) 6= 1,
××then
p there is an isomorphism of GSO(X) with (k × Gl(2, K))/K . Here, K =k( disc(X)). If disc(X) = 1, we have a bijection between Irr(GSO(X)) and thesubset of τ ⊗C τ 0 in Irr(Gl(2, k) × Gl(2, k)) or Irr(D× × D× ) such that ωτ = ωτ 0 . Ifdisc(X) 6= 1, there is a two-to-one map from Irr(GSO(X)) onto the subset of τ in
Irr(Gl(2, K)) such that ωτ factors through NK
k ; the two representations lying over
τ correspond to the characters through which ωτ factors.
Using these identifications, regular, invariant and distinguished have the following meanings for an element π of Irr(GSO(X)). Suppose disc(X) = 1, and letτ ⊗C τ 0 correspond to π. Then π is regular if and only if τ  τ 0 , and if π is invariant, then π is distinguished. Suppose disc(X) 6= 1, and let π correspond to τ andthe quasi-character χ of k × . In this case, π is regular if and only if τ is not Galois
invariant. In contrast to the case disc(X) = 1, not all invariant representations are
distinguished. Indeed, if π is invariant and infinite dimensional, so that τ is Galois
invariant, then π is distinguished if and only if
HomGl(2,k) (τ, χ ◦ det) 6= 0.
Nonvanishing is given by the following theorem. We claim no originality for thisresult. The proof is a straightforward generalization of arguments from [H] and [F],along with some observations from [HST] or [T]. In the global case, this theorem
goes back to [HLR].
Theorem 5.3 (Hakim-Flicker). Let τ in Irr(Gl(2, K)) be infinite dimensional
and Galois invariant. Let ωτ = χ ◦ NK
k . Then the following are equivalent:
1. HomGl(2,k) (τ, χ ◦ det) 6= 0;
2. For every quasi-character ζ of K × extending χ,(τ ⊗C ζ −1 , 1/2, ψK ) = χ(−1);
3. τ is the base change of an element of Irr(Gl(2, k)) with central characterχωK/k .
For explanations of the notation, see section 5.
With these interpretations, we can explain the proof of the main theorem. Under
the hypotheses of the theorem we need to show that every element of the form π −is not in R2 (GO(X)), and that every element of Irr(GO(X)) not of the form π −is in R2 (GO(X)). The first statement follows by an argument analogous to one in[HK]. This proof depends on a lemma that holds more generally: every distribution
on X 2 invariant under SO(Y ) is invariant under O(Y ), for any Y as above.
To prove the second statement, we use the local analogue of the global methodof computing a Fourier coefficient. Let σ in Irr(GO(X)) not be of the form π − . Let

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z be in X 2 . If det(zi , zj ) 6= 0, we will say that z is nondegenerate. As above, if z is
nondegenerate, then the components of z generate a nondegenerate subspace, andthe stabilizer in O(X) is isomorphic to O(Z), where Z is the orthogonal complementof the subspace. By a result analogous to the well known relation between thenonvanishing of global theta lifts and the nonvanishing of period integrals, to showthat σ is in R2 (GO(X)) it suffices to show that
HomO(Z) (σ ∨ , 1) 6= 0,
for some nondegenerate z. See section 6 for a proof and an explanation of theanalogy. First consider the case when σ is not induced from a regular element ofIrr(GSO(X)) or is not of the form π + . Then disc(X) 6= 1, and σ is the extensionof an element of Irr(GSO(X)) corresponding to a τ in Irr(Gl(2, K)) and a quasicharacter χ of k × such that
HomGl(2,k) (τ, χ ◦ det) = 0.
With a proper choice of z and quasi-character ζ of K × extending χ, using the
Kirillov model of τ ∨ , we showL(f ) = Z(ζ −1 , f, 1/2)
is the required linear functional. Here, Z(ζ −1 , f, s) is the zeta function associated to
f in τ ∨ and ζ. In particular, the invariance of L follows from the functional equation
for Z(ζ −1 , f, s). When σ is induced from a regular element π of Irr(GSO(X)) or isof the form π + there is a simplification. In this case, by Theorem 4.4, it suffices toshow that
HomSO(Z) (π ∨ , 1) 6= 0
for some nondegenerate z. When X is isotropic we accomplish this by some Kirillov
model constructions, in part analogous to those of the previous paragraph, andwhen X is anisotropic, we use Tunnell’s work [T].
In combination with some other results, we use the main theorem to prove a caseof a conjecture of S.S. Kudla. To state the conjecture, suppose for the moment that
dimk X is arbitrary. For σ1 in Irr(O(X)), let n(σ1 ) be the smallest integer n suchthat σ1 occurs in the theta correspondence with Sp(n, k).
Conjecture 7.1 (S.S. Kudla). If σ1 is in Irr(O(X)), thenn(σ1 ) + n(σ1 ⊗C sign) = dimk X.
This conjecture is known to be true when dimk X = 0 or 2, but is open for
all other cases. There is another conjecture of S.S. Kudla for representations ofSp(n, k). Recently, S.S. Kudla and S. Rallis have announced considerable progresson this complementary conjecture. See section 7. We prove the following theorem.
Theorem 7.8. Let dimk X = 4 and σ be in Irr(GO(X)). Thenn(σ) + n(σ ⊗C sign) = 4.
Here, n(σ) is defined just as in the case of isometries. To prove the theorem, wecharacterize R1 (GO(X)) and R3 (GO(X)). That is, under the same assumptions asin the theorem, we specify n(σ). To do so, we make use of the main theorem and [S]and [Co]. See Theorem 7.4 and Lemma 7.7. For a presentation of the information,
see the tables in section 7. The following result is an immediate corollary of thetheorem.

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)785Corollary 7.9. If dimk X = 4, then S.S. Kudla’s Conjecture 7.1 is true.
In this paper we do not consider applications to functorality and the theory of
L-packets. For some discussion of these topics see [V] and [HST].
We will now make some remarks about previous work on the Weil representationand theta correspondence for similitudes when dimk X = 4 and n = 2. In [PSS]
and [So1], in the case disc(X) = 1 and X isotropic, the induced Weil representation[R] is used to lift elements of Irr(GSO(X)) to representations of GSp(2, k). This
construction is an analogue of the global definition of theta lifts, and uses elementsof Whittaker models in place of automorphic forms. The problem of whether these
representations of GSp(2, k) are irreducible is not resolved in [PSS] or [So1]. The
work [HPS] in part investigates the case disc(X) = 1 and X anisotropic. In thiscase, as a consequence of Theorem 9.1 of [HPS], every element of Irr(GSO(X)) is an
SO(X) quotient of ω. Using this result, one could prove the main theorem in thiscase using Theorems 4.3 and 4.4. Using the induced Weil representation, resultsfrom the previously mentioned papers, and the strong multiplicity one theorem
for GSp(2) of [So2], a global argument in [V] lifts elements of Irr(GSO(X)) thatare the local components of cuspidal, not invariant, automorphic representationsof GSO(X) to Irr(GSp(2, k)). Included in these representations are the supercuspidal representations. Since it uses Whittaker models, in the case disc(X) 6= 1, thismethod fails to construct the representations that correspond to one of the extensions to GO(X) of the invariant but not distinguished elements of Irr(GSO(X)).
Finally, [HST] makes many remarks and observations about the cases when X isisotropic, though it is mainly concerned with a certain global theta lifting, and its
application to another problem. In particular, after the computation of the Fourier
coefficient of the global theta lift it makes a conjecture essentially equivalent to themain theorem in the case where X is isotropic; see the guess on page 399. However,instead of using the concept of distinguished representations, the guess is phrased
in terms of  factors. Even so, we rely heavily on the understanding of these factors from Lemma 14 of [HST].
In the first section we recall the theory of the theta correspondence for similitudes from [R]. In the second section we characterize GO(X) in terms of the unitsof quaternion algebras. Using this account, in the third section we parameterize
Irr(GO(X)). In the fourth section we define the concept of being distinguished andrelate it to the theta correspondence. Distinguished representations for disc(X) 6= 1
are investigated in the fifth section. The main theorem is proven in the sixth section.
In the remaining section we make the application to S.S. Kudla’s conjecture.
I would like to thank S.S. Kudla for many useful comments, especially for tellingme about his conjectures and the proof of Lemma 4.2. Thanks are also due to J.
Hakim for some helpful conversations concerning his theorem.
We use the following notation. Let J be a group of td-type, as in [C]. Then
Irr(J) is the set of equivalence classes of smooth admissible irreducible representations of J. If π is in Irr(J), then π ∨ in Irr(J) is the contragredient representationof π, and ωπ is the central character of π. A quasi-character of J is a continuous
homomorphism from J to C× , and a unitary character of J is a continuous homomorphism from J to the group of complex numbers of absolute value 1. The
trivial representation of J on C will be denoted by 1. We will also use the notationof [GK] for restriction theory. Throughout the paper k is a nonarchimedean local
field of characteristic zero and odd residual characteristic. Let D be the division

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quaternion algebra over k, with canonical involution ∗ and reduced norm N definedby N (x) = xx∗ = x∗ x. The canonical involution of the quaternion algebra M2 (k)will also be denoted by ∗; in this case the reduced norm is det. We let H denote thehyperbolic plane defined over k. Let ( , )k denote the Hilbert symbol of k. If K is×a quadratic extension of k, then ωK/k is the nontrivial character of k × / NKk (K ).
××2
For d in k /k we let (d) = (−1, −d)k .
1. The theta correspondence for similitudes
In this section we recall some results and definitions from [R]. Suppose that
(X, ( , )) is a nondegenerate symmetric bilinear space over k of even dimension m,
and let n be a nonnegative integer. Let GO(X) be the group of k linear automorphisms h of X such that there exists λ in k × such that (h(x), h(y)) = λ(x, y) forx and y in X. If h is in GO(X), then such a λ is unique, and will be denoted by
λ(h). Let O(X) be the subgroup of all h in GO(X) such that λ(h) = 1. Let sign :GO(X) → {±1} be the unitary character defined by sign(h) = det(h)/λ(h)m/2 .
We let GSO(X) = ker(sign). We will often describe GO(X) in terms of GSO(X)and an extra element of GO(X). Let h0 in GO(X) be such that h20 = 1 and h0
is not in GSO(X). There is an action of the group {1, h0 } on GSO(X) given by
h0 · h = h0 hh0 , and an isomorphism GSO(X) o {1, h0 } ∼= GO(X) that takes (h, δ)to hδ. Next, let GSp(n, k) be the group of all g in Gl(2n, k) such that for some λin k × ,




0
1n
0
1n
t
g=λ
.
g
−1n 0
−1n 0
Again, if g is in GSp(2n, k), then such a λ is unique and will be denoted by λ(g). Let
Sp(n, k) be the subgroup of all g in GSp(n, k) such that λ(g) = 1. Let GSp(n, k)+
be the subgroup of all g in GSp(n, k) such that there exists h in GO(X) such that
λ(g) = λ(h). The group GSp(n, k)+ depends on X and is a proper subgroup ofGSp(n, k) if and only if disc(X) 6= 1. Here, disc(X) is the discriminant of X, as
defined in [Sc]. If disc(X) 6= 1, then [GSp(n, k) : GSp(n, k)+ ] = 2. Fix a nontrivial
additive character ψ of k.
To ψ, X and n, there is associated the Weil representation ω of Sp(n, k) ×
O(X) on S(X n ). For the most part, in this paper we only will need to know theaction of ω(1, h) for h in O(X), which is given by left translation:
ω(1, h) · ϕ(x) = L(h)ϕ(x) = ϕ(h−1 x).
There exists of an extension of ω to a representation of the larger group
R = {(g, h) ∈ GSp(n, k) × GO(X) : λ(g) = λ(h)}.
This extension, called the extended Weil representation, will also be denotedby ω, and is very simply defined by


1
0
− mn
4
, 1)L(h)ϕ.
ω(gω(g, h)ϕ = |λ(h)|0 λ(g)−1
Note that R involves only GSp(n, k)+ . Occasionally, to indicate the dependence ofω on X and n, we will write ωX,n , ωn or ωX for ω.
The Weil representation defines a correspondence between Irr(Sp(n, k)) andIrr(O(X)). Let RX (Sp(n, k)) be the set of all elements of Irr(Sp(n, k)) that areTHE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)787

nonzero quotients of ω, and similarly define Rn (O(X)) . As a consequence of amore general theorem of [W], we have
Theorem 1.1 (Waldspurger). The set{(π, σ) ∈ RX (Sp(n, k)) × Rn (O(X)) : HomSp(n,k)×O(X) (ω, π ⊗C σ) 6= 0}
is the graph of a bijection between RX (Sp(n, k)) and Rn (O(X)).
A correspondence for similitudes is defined by the extended Weil representation.
Let RX (GSp(n, k)+ ) be the set of σ in Irr(GSp(n, k)+ ) such that σ|Sp(n,k) is multiplicity free and has a constituent in RX (Sp(n, k)). Similarly define Rn (GO(X)).
From [R], section 4, we have
Theorem 1.2. The set{(π, σ) ∈ RX (GSp(n, k)+ ) × Rn (GO(X)) : HomR (ω, π ⊗C σ) 6= 0}
is the graph of a bijection between RX (GSp(n, k)+ ) and Rn (GO(X)).
If π is in RX (GSp(n, k)+ ) or σ is in Rn (GO(X)), then we denote the corresponding elements of Rn (GO(X)) and RX (GSp(n, k)+ ) by θ(π) and θ(σ), respectively.
The problem of whether the extended Weil representation defines a well behaved correspondence between Irr(GSp(n, k)) and Irr(GO(X)) when GSp(n, k)+ is
a proper subgroup of GSp(n, k) is also dealt with in [R]. To describe the results,
suppose that GSp(n, k)+ is a proper subgroup of GSp(n, k), i.e., that disc(X) 6= 1.
Then the multiplicity free assumption is unnecessary since [GSp(n, k)+ : k × ·Sp(n, k)] = [GO(X) : k × · O(X)] = 2. See, for example, [GK]. One would liketo know if the condition
HomR (ω, π ⊗C σ) 6= 0
defines a bijection between RX (GSp(n, k)), the set of all π in Irr(GSp(n, k)) suchthat some constituent of π|Sp(n,k) lies in RX (Sp(n, k)), and Rn ((GO(X))). In [R] it
is shown that this condition defines such a bijection if and only if a certain criterion
is satisfied.
To state this criterion, we need to introduce the other nondegenerate symmetricbilinear space X 0 of dimension m and discriminant disc(X). From the Witt decomposition theorem we see that X 0 can be taken to have the same vector space as X,
but with symmetric bilinear form multiplied by an element of k × . Assume that X 0
has this form. Then GO(X) = GO(X 0 ), and the restrictions of the Weil representations ω and ω 0 associated to X and X 0 , respectively, to O(X) = O(X 0 ) are identical.
It follows that Rn (O(X)) = Rn (O(X 0 )) and Rn (GO(X)) = Rn (GO(X 0 )). However,the correspondences defined by ω and ω 0 may differ. In [R] it is proven that theabove condition defines a bijection if and only if the correspondences defined by ωand ω 0 are disjoint, i.e., RX (Sp(n, k)) ∩ RX 0 (Sp(n, k)) = ∅.
Suppose RX (Sp(n, k)) ∩ RX 0 (Sp(n, k)) = ∅. From [R] we have the following. Let
g be a representative for the nontrivial coset of GSp(n, k)/ GSp(n, k)+ . Let σ bein Rn (GO(X)) = Rn (GO(X 0 )), and let π and π 0 in Irr(GSp(n, k)+ ) correspond to
σ with respect to ω and ω 0 , respectively. Then g · π = π 0 , andGSp(n,k)

IndGSp(n,k)+ π
in RX (GSp(n, k)) corresponds to σ.

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Whether the criterion is expected to hold depends on m and n. If the underlying
bilinear spaces lie in the stable range, i.e., if m ≥ 4n + 2, then the criterion does
not hold. From [HKS], we have have the following conjecture.
Conjecture 1.3 (Theta dichotomy). If m ≤ 2n, thenRX (Sp(n, k)) ∩ RX 0 (Sp(n, k)) = ∅.
For progress on the conjecture, see [KR] and [HKS]. The theta dichotomy conjecture follows from another strong and precise conjecture of S.S. Kudla. Recently,
S.S. Kudla and S. Rallis have announced considerable progress on this stronger
conjecture. In particular, it follows from their result that no supercuspidal representation is contained in the intersection of Conjecture 1.3. See section 7.
2. Four dimensional symmetric bilinear spaces
In this section we recall the characterization of the group of similitudes of a four
dimensional symmetric bilinear space in terms of the units of a quaternion algebra.
××2
For the remainder of this paper, d will
√ be an element of k /k . If d = 1, then let
K = k × k; if d 6= 1, then let K = k( d). Let Gal(K/k) = {1, −}.
Let B be a quaternion algebra over K, with canonical involution ∗. A k linear
ring automorphism s of B is a Galois action on B if s2 = 1 and s(ax) = as(x)for a in K and x in B. Let s be a Galois action on B. Let X(s) be the set of all
x in B such that s(x) = x∗ . Then X(s) is a four dimensional vector space overk, and equipped with the restriction of the symmetric bilinear form corresponding
to the reduced norm of B, X(s) is a nondegenerate symmetric bilinear space. The
discriminant and Hasse invariant of X(s) are d and (d)(s), respectively. Here, todefine (s), let B(s) be the fixed points of s. Then B(s) is a quaternion algebra
over k, and (s) = 1 if B(s) is split and (s) = −1 if B(s) is ramified.
The elements of k × × B × give elements GSO(X). Define a left action ρ of×
k × B × on X(s) by ρ(t, g)x = t−1 gxs(g)∗ . Then ρ(t, g) is in GSO(X(s)) for (t, g)
in k × × B × . There is an inclusion of K × in k × × B × that sends a to (NKk (a), a).
The following result is well known.
Theorem 2.1. For every four dimensional nondegenerate symmetric bilinear space
X of discriminant d over k there exists a quaternion algebra B over K and a
Galois action s on B such that X ∼= X(s) as symmetric bilinear spaces. For every
quaternion algebra B over K and Galois action s on B the sequenceρ

1 → K × → k× × B × −→ GSO(X(s)) → 1
is exact.
We now define concrete realizations of the two four dimensional nondegenerate
symmetric bilinear spaces X(d, ) of discriminant d and Hasse invariant  in {±1}.
Suppose first d = 1. Let B be M2 (k) × M2 (k) or D × D. Define a Galois action on
B by s(x, y) = (y, x). Then X(s) is isomorphic to M2 (k) or D. We find that M2 (k)and D have discriminant 1 and Hasse invariant (d) and −(d), respectively. We
let X(1, (d)) = M(k) and X(1, −(d)) = D. The above exact sequence simplifies
toρ→ GSO(X(1, (1))) → 1
1 → k × → Gl(2, k) × Gl(2, k) −THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)

789

and
ρ

→ GSO(X(1, −(1))) → 1,
1 → k× → D× × D× −where ρ is now defined by ρ(g, g 0 )x = gxg 0∗ , and the inclusion of k × sends x to(x, x−1 ).
Suppose that d 6= 1. Let B = M2 (K). Then B = K ⊗k M2 (k) or B = K ⊗k D.
Here we regard D as a subalgebra of B by lettinga bδ
: a, b ∈ K}
D={
b a
×
where δ is a representative for the nontrivial coset of k × / NKk (K ). Define Galois
0
actions s and s on B corresponding to the decompositions B = K ⊗k M2 (k) and
B = K ⊗k D, respectively, by s(a ⊗ x) = a ⊗ x and s(a ⊗ y) = a ⊗ y for a in K,
x in M2 (k) and y in D. X(s) and X(s0 ) have discriminant d and Hasse invariants
(d) and −(d), respectively. We let X(d, (d)) = X(s) and X(d, −(d)) = X(s0 ).
There are exact sequencesρ→ GSO(X(d, ±(d))) → 1.
1 → K × → k × × Gl(2, K) −Explicitly,

√ 
a
b d
√
: a ∈ K, b, c ∈ k}X(d, (d)) = {
a
c d


and


X(d, −(d)) = {

b
a


−δa
: a ∈ K, b, c ∈ k}.
c

As mentioned in section 1, there is a similitude between X(d, (d)) and X(d, −(d)).
Let us construct this similitude explicitly. Since s◦s0 is a K algebra isomorphism, by
the Skolem-Noether theorem there exists a u in Gl(2, K) such that s0 (x) = us(x)u−1
for x in M2 (K). We may assume that us(u) = us0 (u) = det(u). Then the map
S : X(d, (d)) → X(d, −(d)) defined by S(x) = xu−1 is a well defined similitude
such that S(ρ(t, g)x) = ρ(t, g)S(x) for (t, g) in k × × Gl(2, K) and x in X(d, (d)).
For the remainder of this paper,  will be in {±1}, and X = X(d, ). Also, h0
is the element of GO(X) that sends x to x∗ . Because of the remarks in section 1
concerning the theta correspondence for similitudes when d 6= 1, we will disregard
the case d 6= 1 and  = −(d). Thus, if d 6= 1, then X = X(d, (d)). We will letω denote the extended Weil representation associated to X and the nonnegative
integer n. If necessary, the dependence of ω on n will be indicated by a subscript.
3. Representations
In this section we make some definitions and elementary observations concerning
the relationship between representations of GO(X) and GSO(X) and the quaternion algebras from the last section. We remind the reader that the case d 6= 1
and  = −(d) for our purposes can be and will be ignored. We also point out
that by [HPS], Lemma 7.2, the restriction of representations of GO(X) to O(X) is
multiplicity free.
Suppose first that d = 1. Let Irrf (Gl(2, k) × Gl(2, k)) be the set of pairsof representations in Irr(Gl(2, k)) with the same central character. We define

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Irrf (D× × D× ) similarly. There are bijections∼

Irr(GSO(X(1, (1)))) −→ Irrf (Gl(2, k) × Gl(2, k))and∼Irr(GSO(X(1, −(1)))) −→ Irrf (D× × D× )that take π to the representation that sends (g, g 0 ) to π(ρ(g, g 0 )). If (τ, τ 0 ) is contained in Irrf (Gl(2, k)×Gl(2, k)) or Irrf (D× ×D× ), then the corresponding element
π(τ, τ 0 ) of Irr(GSO(X(1, ±(d)))) has as space the space of τ ⊗C τ 0 and is definedby π(τ, τ 0 )(ρ(g, g 0 )) = τ (g) ⊗ τ 0 (g 0 ). The central character of π(τ, τ 0 ) is ωτ = ωτ 0 ,and the contragredient of π(τ, τ 0 ) is π(τ, τ 0 )∨ = π(τ ∨ , τ 0∨ ).
Suppose that d 6= 1. Let Irrf (Gl(2, K)) be the set of elements of Irr(Gl(2, K))
with Galois invariant central character. Recall that if a quasi-character of K × is×Galois invariant, then it factors through NK
k via exactly two quasi-characters of k .
There is a two-to-one surjective map
Irr(GSO(X(d, (d)))) → Irrf (Gl(2, K))
that takes π to the representation that has space the space of π and is defined by
g 7→ π(ρ(1, g)). If τ is in Irr(Gl(2, K)), and χ and χ0 are the two quasi-charactersK0of k × such that ωτ = χ ◦ NK
k and ωτ = χ ◦ Nk , then the two elements π(τ, χ)
0and π(τ, χ ) of Irr(GSO(X(d, (d)))) lying over τ are defined by π(τ, χ)(ρ(t, g)) =χ(t)−1 τ (g) and π(τ, χ0 )(ρ(t, g)) = χ0 (t)−1 τ (g). The central character of π(τ, χ) isχ, and the contragredient of π(τ, χ) is π(τ, χ)∨ = π(τ ∨ , χ−1 ).
Having described the representations of GSO(X), we consider their relationshipto representations of GO(X). Let π be in Irr(GSO(X)). If the induced representation of π to GO(X) is irreducible, we say that π is regular, and if the induced
representation of π to GO(X) is reducible, we say that π is invariant. If π isregular, we denote the induced representation of π to GO(X) by π + .
We can describe regular and invariant representations in terms of the above
characterizations.
Proposition 3.1. Let π be in Irr(GSO(X)). If d = 1, then π is invariant if andonly if π = π(τ, τ ) for some τ in Irr(Gl(2, k)) or Irr(B × ). If d 6= 1, then π isinvariant if and only if π = π(τ, χ) for some Galois invariant τ in Irr(Gl(2, K)).
It also will be useful to have an explicit description of the finite dimensional
elements of Irr(GSO(X)) and Irr(GO(X)) in the case d 6= 1. Assume d 6= 1. The
finite dimensional elements of Irr(GSO(X)) are one dimensional, and of the form
π(β, χ), where β and χ are quasi-characters of K × and k × , respectively, such thatβ 2 = χ ◦ NK
k . Here, to avoid excessively complicated notation, in π(β, χ) we regard
β as the quasi-character of Gl(2, K) that sends g in Gl(2, K) to β(det(g)). It follows
that the finite dimensional elements of Irr(GO(X)) are either one or two dimensional. One dimensional elements arise from invariant elements of Irr(GSO(X)),
while two dimensional elements come from regular elements of Irr(GSO(X)).
Proposition 3.2. Suppose d 6= 1. The subset of invariant one dimensional elements of Irr(GSO(X)) consists of the π(α ◦ NK
k , χ), where α and χ are quasiK2characters of k × and χ ◦ NK
k = α ◦ Nk . The set of finite dimensional regularelements of Irr(GSO(X)) consists of the π(β, χ), where β is a Galois noninvariant
quasi-character of K × and χ is a quasi-character of k × such that β 2 = χ ◦ NK
k .

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Moreover, if π(β, χ) is such a regular element of Irr(GSO(X)), then χ = β|k× µ forsome nontrivial quadratic quasi-character µ of k × different from ωK/k .
4. Distinguished representations and the correspondence
In this section we will define what it means for an invariant representation of
GSO(X) to be distinguished, and we will consider what effect being distinguishedhas on what extensions of the representation to GO(X) occur in the theta correspondence. The idea that certain extensions of a distinguished representation
cannot occur in the theta correspondence is due to [HK]. This appears in Theorem
4.3 below. We go a step further, and show how an extension of a distinguishedrepresentation can be proven to occur in the theta correspondence. See Theorem
4.4.
Let π be in Irr(GSO(X)). To define what it means for π to be distinguished,suppose y in X is anisotropic. Then the stabilizer in O(X) of y can be identified
with O(Y ), where Y is the orthogonal complement to y, and we will write O(Y )for this stabilizer. We say that π is generically distinguished if π is invariant,and there is an anisotropic y in X such that
HomSO(Y ) (π, 1) 6= 0,
and, if d 6= 1, then Y is isotropic. We will see in Corollary 7.5 that the assumptionthat Y be isotropic when d 6= 1 is unnecessary, but it is convenient to make it. We
will say that π is distinguished if π is generically distinguished or d 6= 1 and π isinvariant and one dimensional.
We regard distinguished, but not generically distinguished, representations asboundary cases. Our reason for including them is, of course, that they behave likegenerically distinguished representations with respect to the theta correspondence.
We also note that they are irreducible subquotients of generically distinguishedreducible principal series representations.
In fact, a representation is generically distinguished if and only if it is generically distinguished with respect to a certain anisotropic y0 in X. Define y0 in thefollowing way. If d = 1, let y0 = 1. If d 6= 1, also let y0 = 1. Using the Wittcancellation theorem and the Witt extension theorem, one can show that if y is asin the last paragraph, then there exists h in GSO(X) such that h(y) = y0 . It follows that a representation is generically distinguished if and only if it is genericallydistinguished with respect to y0 . For the remainder of this paper we let Y be theorthogonal complement to y0 .
The group SO(Y ) can be concretely described. If d = 1, then SO(Y ) is theimage under ρ of the subgroup {(g, g ∗−1 ) : g ∈ Gl(2, k)} or {(g, g ∗−1 ) : g ∈ D× }. If
d 6= 1, then by Hilbert’s Theorem 90, SO(Y ) is the image under ρ of the subgroup{(det(g), g) : g ∈ Gl(2, k)}. We also note that h0 fixes y0 , and thus is contained inO(Y ). Together, SO(Y ) and h0 generate O(Y ).
In the case d = 1, the next proposition completely identifies all the distinguishedrepresentations. We will consider the case d 6= 1 in greater detail in the next section.
Proposition 4.1 (Hakim). Let π ∈ Irr(GSO(X)). Assume that π is invariant.
Then
dimC HomSO(Y ) (π, 1) ≤ 1.

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If d = 1, then π is generically distinguished. If d 6= 1 and π = π(τ, χ), then π is
generically distinguished if and only if
HomGl(2,k) (τ, χ ◦ det) 6= 0.
Proof. Suppose that d = 1. Since π is invariant, it follows that π = π(τ, τ ) for some
τ in Irr(Gl(2, k)) or τ in Irr(D× ). Now τ ∨ ∼= ωτ−1 ⊗C τ . It follows that there is an
isomorphism
HomSO(Y ) (π, 1) ∼= HomGl(2,k) (τ ⊗C τ ∨ , 1)
or an isomorphism
HomSO(Y ) (π, 1) ∼= HomD× (τ ⊗C τ ∨ , 1).
Here Gl(2, k) or D× is embedded on the diagonal. It is well known that the second
homomorphism space has dimension one.
Suppose that d 6= 1. Then there is an isomorphism
HomSO(Y ) (π, 1) ∼= HomGl(2,k) (τ, χ ◦ det).
By an argument as in [H], this space has dimension less than or equal to 1.
If a representation is distinguished, then we will identify its extensions to GO(X)in the following way. Suppose first that π in Irr(GSO(X)) is generically distinguished. Since by Proposition 4.1
dimC HomSO(Y ) (π, 1) = 1,
it follows that for exactly one extension π 0 of π to GO(X) we have HomO(Y ) (π 0 , 1) 6=
0. Call this extension π + ; the other extension of π to GO(X) will be called π − .
From our above remarks, the definitions of π + and π − do not depend on the choiceof y. Suppose next that d 6= 1 and π = π(α◦NKk , χ) is invariant and one dimensional.
See Proposition 3.2. We will denote the extension of π that sends h0 to χ(−1) by
π + ; the other extension of π will be denoted by π − . It can be verified that the two
definitions of π + and π − in the case where d 6= 1 and π is generically distinguishedand one dimensional are the same.
The following lemma will be essential in determining which extensions of a generically distinguished representation occur in the theta correspondence. As we mentioned earlier, the proof of this lemma is general. A form of a case of this lemma
appears in Proposition 4.5 of [P1].
Lemma 4.2. If n < dimk Y , i.e., if n = 1 or 2, then any distribution on X ninvariant under SO(Y ) is invariant under O(Y ).
Proof. Assume n < dimk Y . It is straightforward to see that any distribution on X ninvariant under SO(Y ) is invariant under O(Y ) if and only if HomO(Y ) (ωX,n , sign) =0. Assume that HomO(Y ) (ωX,n , sign) 6= 0. Since X = Y ⊥ Y ⊥ , we have ωX,n ∼=ωY,n ⊗ ωY ⊥ ,n , as O(Y ) × O(Y ⊥ ) representations. Since HomO(Y ) (ωX,n , sign) 6= 0,
it follows that HomO(Y ) (ωY,n , sign) 6= 0. This contradicts the appendix of [Ra].
The next theorem shows that at most one of the extensions of a genericallydistinguished representation can occur in the theta correspondence when n is 1 or
2. As we pointed out above the idea is due to [HK]. In [HK] the case n = 1 wasconsidered. We shall see in Lemma 6.5 that the result holds for all distinguishedrepresentations, not just for generically distinguished representations.

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)793

Theorem 4.3. Let π ∈ Irr(GSO(X)). If π is generically distinguished, then π − isnot in Rn (GO(X)) for n < dimk Y , i.e., for n = 1 and 2.
Proof. We begin with two comments concerning π. First, π|SO(X) is multiplicity
free. For let
π|SO(X) = m · π1 ⊕ · · · ⊕ m · πM ,where the πi ∈ Irr(SO(X)) are mutually inequivalent, and m and M are positive
integers. Then
M
X

m · dimC HomSO(Y ) (πi , 1) = 1,i=1which implies that m = 1, and that dimC HomSO(Y ) (πi , 1) = 1 for exactly one i, say
i = 1, and dimC HomSO(Y ) (πi , 1) = 0 for i > 1. Second, suppose that Vi is the spaceof πi ; we assert that π + (h0 )Vi = Vi for all i. Let us prove this first when i = 1. Let
π + (h0 )V1 = Vi . Let f in HomSO(Y ) (π1 , 1) be nonzero. Define a linear functional f 0
on Vi by f 0 (v) = f (π + (h0 )v). Then f 0 is in HomSO(Y ) (πi , 1). Since f 0 6= 0, i = 1.
Let i be arbitrary. There exists h in GSO(X) such that π(h)V1 = Vi . We have
π + (h0 )Vi = π + (h0 h)V1 = π + (h0 hh0 )V1 = π(h)π(h−1 h0 hh0 )V1 = π(h)V1 = Vi ,since h−1 h0 hh0 is in SO(X).
Suppose that π − is in Rn (GO(X)) for n = 1 or 2. Then there exists a nonzero
O(X) map T from ωn to π − . Let f ∈ HomSO(Y ) (π, 1) be nonzero. We may assume
the composition f ◦ T is nonzero. This is a nonzero SO(Y ) invariant distribution on
X n . By Lemma 4.2, f ◦ T is invariant under h0 . But since T is an O(X) map andby the definition of π − , the composition of h0 with f ◦ T is −f ◦ T . Since f ◦ T 6= 0,
this is a contradiction.
The next theorem gives a sufficient condition for one of the extensions of a generically distinguished representation to occur in the theta correspondence. Again, weshall see in Lemma 6.5 that the result holds for all distinguished representations,
not just for generically distinguished representations.
Theorem 4.4 . Let π ∈ Irr(GSO(X)). Suppose π is regular or generically distinguished, and n < dimk Y , i.e., n = 1 or 2. ThenHomSO(X) (ωn , π) 6= 0 =⇒ HomO(X) (ωn , π + ) 6= 0.
Proof. Suppose first that π is regular. Let V be the space of π. As a model for π + we
can take the representation with space V ⊕ V and action defined by π + (h)(v ⊕ v 0 ) =π(h)v ⊕ π(h0 hh0 )v 0 for h in GSO(X) and π + (h0 )(v ⊕ v 0 ) = v 0 ⊕ v. Let L in
HomSO(X) (ωn , π) be nonzero. Define L0 : ωn → π + by L0 (ϕ) = L(ϕ) ⊕ L(ωn (h0 )ϕ).
Then L0 is in HomO(X) (ωn , π + ) and L0 is nonzero.
Suppose that π is generically distinguished. We will use the notation of the proofof Theorem 4.3. Let L in HomSO(X) (ωn , π) be nonzero. We may assume that thecomposition L1 of L with the projection of V onto V1 is nonzero. To complete
the proof it suffices to show that L1 ◦ ωn (h0 ) = π + (h0 ) ◦ L1 . We first show thatωn (h0 ) ker(L1 ) = ker(L1 ). Suppose not, i.e., suppose that L1 (ωn (h0 ) ker(L1 )) 6= 0.
Then by the irreducibility of π1 , L1 (ωn (h0 ) ker(L1 )) = V1 . Let f in HomSO(Y ) (π1 , 1)
be nonzero. Consider f ◦ L1 . This distribution is nonzero and SO(Y ) invariant. By
Lemma 4.2, f ◦ L1 is invariant under h0 , so that f (V1 ) = f (L1 (ωn (h0 ) ker(L1 ))) =f (L1 (ker(L1 ))) = 0, contradicting f 6= 0. Now since ker(L1 ) is invariant under

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ωn (h0 ), it follows that S(X n )/ ker(L1 ) is an O(X) space. Via the SO(X) isomorphism given by L1 between S(X n )/ ker(L1 ) and V1 we can define an action of h0
on V1 so that L1 is an O(X) map. By Theorem 4.3, this extension must be π + .
A similar argument proves the following statement. Let π be in Irr(GSO(X))
and Π be in Irr(GSp(n, k)+ ), for n = 1 or 2. Assume that π is regular or genericallydistinguished. Then
HomR0 (ωn , Π ⊗C π) 6= 0 =⇒ HomR (ωn , Π ⊗C π + ) 6= 0.
Here R0 is the subset of elements of R whose first entries are in GSO(X).
This result has some interesting consequences. It implies that if a regular or
generically distinguished element of Irr(GSO(X)) corresponds to an element ofIrr(GSp(n, k)+ ), in the obvious sense, then that element of Irr(GSp(n, k)+ ) is
unique. In particular, since all elements of Irr(GSO(X)) are either regular or generically distinguished when disc(X) = 1, it follows that in this case if Π is as above,
then Π is always uniquely determined. When disc(X) = 1 and n = 1 this helps
one to understand the Jacquet-Langlands correspondence from the point of viewof the theta correspondence. See section 7 and [S]. When disc(X) = 1 and n = 2,
using the relation to the alternate approach to similitudes via the induced Weil representation [R], this gives a different argument for part of the proof of the strong
multiplicity one theorem for GSp(2) as in [So2]. It would be interesting to see if
a complete proof could be obtained along these lines. This would require that theresults of this section be extended to the case when X is the split six dimensional
space. To do so, it would seem to be necessary to use a Y with dimk Y = 4. We
also mention that the above development is used implicitly in the classical casedimk X = 2n = 2. In this case one uses a Y with dimk Y = 2, i.e., Y = X . For
more remarks about this case, see section 7.
Finally, the results of this section generalize considerably. To generalize thedefinitions, suppose that X is an even dimensional nondegenerate symmetric bilinear space, and n is a positive integer. In place of the one dimensional subspacek · y above, one could take any nondegenerate subspace, and again define Y as
the orthogonal complement to this subspace. The definition of a generically distinguished representation would be as above. Then the results generalize in thefollowing way. Lemma 4.2 and its proof hold for n < dimk Y . If one assumes thedimension statement of Proposition 4.1, then Theorem 4.3 and Theorem 4.4 also
hold for n < dimk Y . Thus, the remaining key issue for generalization is the validityof the dimension statement of Proposition 4.1. If dimk Y is small with respect todimk X, then we cannot expect the dimension statement to hold in general.
5. Distinguished Gl(2, K) representations
In the last section we reduced the problem of determining the distinguishedrepresentations of GSO(X) in the case d 6= 1 to a problem concerning the corresponding representations of Gl(2, K). The problem of determining distinguishedGl(2, K) representations has essentially been solved by several authors. See [H]and [F]. Ultimately, the consideration of distinguished Gl(2, K) representations
goes back to a global result of [HLR]. However, since a complete account does notappear in the literature we need to give an exposition.
We begin by defining some notation and recalling some facts. Essentially, we will
follow [G]. In this section we assume that d 6= 1 so that K is a quadratic extension

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of k. Let πK be a uniformizer for K, and let ψK be a nontrivial Galois invariant
additive character of K. If τ in Irr(Gl(2, K)) is infinite dimensional, let K(τ, ψK )be the Kirillov model of τ with respect to ψK . Let τ in Irr(Gl(2, K)) be infinite
dimensional. For g in Gl(2, K), ζ a quasi-character of K × , f in K(τ, ψK ), and s inC, letZ
τ (g)f (x)ζ(x)|x|s−1/2 dx.
Z(g, ζ, f, s) =K×

This integral converges absolutely if <(s) is sufficiently large. Moreover, the function defined by the integral for sufficiently large <(s) has an analytic continuation toa meromorphic function on C. There exists a meromorphic function γ(τ ⊗C ζ, s, ψK )on the complex plane such that
0 1g, ζ −1 ωτ−1 , f, 1 − s)γ(τ ⊗C ζ, s, ψK )Z(g, ζ, f, s) = Z(
−1 0
for g in Gl(2, K) and f in K(τ, ψK ). Let
(τ ⊗C ζ, s, ψK ) = γ(τ ⊗C ζ, s, ψK )L(τ ⊗C ζ, s).
L(τ ⊗C ωτ−1 ζ, 1 − s)Here the L factors are as in [G]. The function (τ ⊗C ζ, s, ψK ) is entire, and has
no zeros. The notation for irreducible principal series and special representationsof Gl(2, K) will be as in [GL]. Let π(µ1 , µ2 ) be a principal series representationof Gl(2, K). Then π(µ1 , µ2 ) is Galois invariant if and only if µ1 ◦ − = µ1 andµ2 ◦ − = µ2 , or µ1 ◦ − = µ2 . Let σ(µ1 , µ2 ) be a special representation. Thenσ(µ1 , µ2 ) is Galois invariant if and only if µ1 ◦ − = µ1 and µ2 ◦ − = µ2 .
Lemma 5.1. Let τ in Irr(Gl(2, K)) be infinite dimensional and Galois invariant.
×
whose restriction to k × isLet ωτ = χ ◦ NK
k and let ζ be a quasi-character of K
χ. If τ is not a principal series representation π(µ1 , µ2 ) with µ1 and µ2 Galois
invariant, then the integral Z(g, ζ −1 , f, 1/2) is absolutely convergent for all g in
Gl(2, K) and f in K(τ, ψK ).
Proof. The claim follows if τ is supercuspidal. Assume that τ is a principal series
representation. Then τ = π(µ1 , µ2 ) with µ1 ◦ − = µ2 . It suffices to show that forf ∈ S(K) the integralZ|x|1/2 µ1 (x)f (x)ζ(x)−1 d× x
K×is absolutely convergent. An estimate shows that this integral converges absolutelyif|µ1 (πK )ζ(πK )−1 | < |πK |−1/2 .
Since |µ1 (πK )ζ(πK )−1 |2 = 1, our claim follows. Suppose that τ is a special representation. Then τ = σ(µ1 , µ2 ) with µ1 ◦ − = µ1 , µ2 ◦ − = µ2 and µ1 = µ2 | |.
Again, it suffices to show that the above integral is absolutely convergent. We have|µ1 (πK )ζ(πK )−1 | = |πK |1/2 < |πK |−1/2 .
Lemma 5.2. Let τ , χ and ζ be as in the last lemma. Then γ(τ ⊗C ζ −1 , s, ψK ) isdefined at 1/2 andγ(τ ⊗C ζ −1 , 1/2, ψK ) = (τ ⊗C ζ −1 , 1/2, ψK ).

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Proof. By definition,
(τ ⊗C ζ −1 , s, ψK ) = γ(τ ⊗C ζ −1 , s, ψK )L(τ ⊗C ζ −1 , s).
L(τ ⊗C ωτ−1 ζ, 1 − s)Since (τ ⊗C ζ −1 , s, ψK ) is an entire function, and since the L functions are definedat 1/2 by Lemma 5.1, it suffices to show thatL(τ ⊗C ζ −1 , 1/2)
= 1.
L(τ ⊗C ωτ−1 ζ, 1/2)
If τ is supercuspidal this is clear. Suppose that τ is a principal series representation
π(µ1 , µ2 ) with µ1 ◦ − = µ2 . Then
L(µ1 ζ −1 , 1/2)L(µ2 ζ −1 , 1/2)
L(τ ⊗C ζ −1 , 1/2)
=
.
−1
−1
L(τ ⊗C ωτ ζ, 1/2)L(µ−11 ζ, 1/2)L(µ2 ζ, 1/2)
It will suffice to show that µ1 (πK )2 = ζ(πK )2 if µ1 ζ −1 is unramified and µ2 (πK )2 =ζ(πK )2 if µ2 ζ −1 is unramified. By symmetry, it is enough to prove one of thesestatements. Suppose µ1 ζ −1 is unramified. If K/k is unramified, then this follows
since we can take πK in k × , and µ1 µ2 = ζζ ◦ − and µ1 ◦ − = µ2 . Suppose thatK/k is ramified. Since the residual characteristic of k is odd, we can assume that2πK = −πK and πK
is a uniformizer of k. Then µ1 (πK )2 = µ1 (−1)µ1 (πK )µ2 (πK ) =µ1 (−1)ζ(πK )ζ(πK ) = µ1 (−1)ζ(−1)ζ(πK )2 = ζ(πK )2 , since ζ(−1) = µ1 (−1) because µ1 ζ −1 is unramified. The case when τ is a special representation is analogous;
for details, see the similar case treated in the remark below.
The last lemma does not hold for all irreducible principal series representationsπ(µ1 , µ2 ) with µ1 and µ2 Galois invariant. Indeed, we claim that if τ = π(µ1 , µ2 ) is
an irreducible principal series representation with µ1 and µ2 Galois invariant, thenγ(τ ⊗C ζ −1 , 1/2, ψK ) = (τ ⊗C ζ −1 , 1/2, ψK )unless µ1 ζ −1 is unramified and µ1 (πK )ζ(πK )−1 = |πK |−1/2 or µ2 ζ −1 is unramified
and µ2 (πK )ζ(πK )−1 = |πK |−1/2 ; in these last cases,γ(τ ⊗C ζ −1 , 1/2, ψK ) = −(τ ⊗C ζ −1 , 1/2, ψK ).
To prove these claims we proceed as in the proof of Lemma 5.2. We need to computeL(τ ⊗C ζ −1 , s)L(µ1 ζ −1 , s)L(µ2 ζ −1 , s)
=
lim
.
−1
s→1/2 L(τ ⊗C ωτ−1 ζ, 1 − s)s→1/2 L(µ−11 ζ, 1 − s)L(µ2 ζ, 1 − s)lim

We first show that µ1 ζ −1 is unramified if and only if µ2 ζ −1 is. Suppose that µ1 ζ −1 isKKunramified. Then ζ(u) = µ1 (u) for all u ∈ O×
K . Since µ1 (ker(Nk )) = 1 and ker(Nk )K
2
is contained in O×
K , ζ(ker(Nk )) = 1. So, ζ ◦ − = ζ. Now µ1 µ2 = ζζ ◦ − = ζ .
−1
−1 −1
−1
Hence, µ1 ζ = (µ2 ζ ) , and µ2 ζ is unramified. The converse follows by symmetry. Note that we also have shown that if µ1 ζ −1 and µ2 ζ −1 are unramified, thenµ1 (πK )µ2 (πK ) = ζ(πK )2 , i.e., µ1 (πK )ζ(πK )−1 = µ2 (πK )−1 ζ(πK ). If now µ1 ζ −1
and µ2 ζ −1 are ramified or µ1 ζ −1 and µ2 ζ −1 are unramified and µ1 (πK )ζ(πK )−1 6=|πK |−1/2 and µ2 (πK )ζ(πK )−1 6= |πK |−1/2 , then the limit is 1. Suppose µ1 ζ −1
and µ2 ζ −1 are unramified and µ1 (πK )ζ(πK )−1 = |πK |−1/2 or µ2 (πK )ζ(πK )−1 =|πK |−1/2 . Then exactly one of µ1 (πK )ζ(πK )−1 and µ2 (πK )ζ(πK )−1 is |πK |−1/2 .

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)797

Without loss of generality, we may assume that µ1 (πK )ζ(πK )−1 = |πK |−1/2 . Then
L(µ1 ζ −1 , s)L(µ2 ζ −1 , s)L(µ1 ζ −1 , s)L(µ2 ζ −1 , s)
lim
= lim
−1
−1
−1
, 1 − s) s→1/2 L(µ2 ζ −1 , 1 − s)s→1/2 L(µ1 ζ, 1 − s)L(µ2 ζ, 1 − s)s→1/2 L(µ1 ζ=(−1) · 1 = −1.
lim

The proof of Theorem 5.3 that we now give follows essentially from [H] and from[T], as interpreted in [HST]. The previous discussion shows that in Theorem 5.3 itis essential to use  instead of γ factors. Note also that ψK differs from the additive
character in [H]. There it is assumed that ψK is trivial on k.
Proof of Theorem 5.3. Assume first that τ 6= π(µ1 , µ2 ) with µ1 and µ2 Galois invariant.
(1) ⇐⇒ (2): The equivalence follows from Lemmas 5.1 and 5.2 and an argument
essentially as in the proof of Theorem 4.1 of [H].
(2) ⇐⇒ (3): Since τ is Galois invariant τ and τ 6= π(µ1 , µ2 ) with µ1 and µ2
Galois invariant, τ is the base change of a discrete series representation of Gl(2, k)that has central character χ or χωK/k . The equivalence of (2) and (3) is 4 of Lemma
14 of [HST].
Now suppose that τ = π(µ1 , µ2 ) with µ1 and µ2 Galois invariant. We will showthat (1), (2) and (3) all hold. The statement (2) follows from Lemma 14 of [HST].
00
To see (3), note that µ1 and µ2 factor through NK
k via, say, µ1 and µ2 , respectively.
00
0 0
By replacing µ1 by ωK/k µ1 , if necessary, we may assume that µ1 µ2 = χ. Since±1
0 0 −100
6= | |±1µ1 µ−1
2 6= | |K it follows that µ1 µ2
k . It follows that π(µ1 , µ2 ) is defined,
00
and the base change of π(µ1 , µ2 ) is τ . To show (1), we proceed as in Proposition 9of [F]. Let
√ 
 √
d
− d
g0 =
1
1
and



a bd
T ={
: a, b ∈ k, a2 − db2 6= 0},b aThen g0 T 0 g0−1 = T andg0−1 Gl(2, k)g0 = {Define L : π(µ1 , µ2 ) → C byZ

L(f ) =
T \ Gl(2,k)


a
b

b
a



z 0
T ={
: z ∈ K × }.
0 z
0: aa − bb 6= 0}.

f (g0−1 g)χ(det(g))−1 dg.

A computation shows that the integrand is well defined. Moreover, one can showthat T \ Gl(2, k) has finite measure and that the integrand is bounded, so that the
integral converges. Finally, L is nonzero and contained in HomGl(2,k) (τ, χ◦det).
6. The main theorem
In this section we prove the main theorem. Our method for showing that a
representation occurs in the theta correspondence is entirely analogous to the global
technique of computing a Fourier coefficient of a global theta lift. To explain the

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BROOKS ROBERTS

analogy, we first give some definitions. Let Nn be the unipotent radical of the Siegel
parabolic in Sp(n, k), i.e., the elements of the form


1 b
0 1
for b in Mn (k) with t b = b. Given β in Mn (k) with t β = β we can define a characterψβ of Nn by


1
1 b
= ψ( tr(bβ)).
ψβ0 12Here, ψ is our fixed nontrivial character of k. Let u be in X n . We say that u isnondegenerate if the components of u generate a nondegenerate subspace of X,or, equivalently, if det(ui , uj ) 6= 0. Let u be nondegenerate. The stabilizer of u in
O(X) can be identified with O(U ), where U is the orthogonal complement to thesubspace generated by the components of u.
With this notation we can explain our method and its analogy to the global
computation of a Fourier coefficient. Suppose for the moment k is a number field,
and σ1 is an irreducible cuspidal automorphic representation of O(X(A)). Let f
be in σ1 , let ϕ be in S(X(A)n ), and let F be the theta lift of f to Sp(n, A) withrespect to ϕ. Let u in X(k)n be nondegenerate, and set β = (ui , uj ). The Fourier
coefficient of F with respect to β is the functionZ
F (ng)ψβ (n) dn.
Fβ (g) =Nn (k)\Nn (A)A computation shows that this isZZωn (g, h)ϕ(u)O(U(A))\ O(X(A))

f (h0 h) dh0 dh.
O(U (k))\ O(U (A))

In conclusion, we find that the theta lift Θ(σ1 ) of σ1 to Sp(n, A) has a nonzero
Fourier coefficient with respect to β if and only ifZf (h) dh 6= 0
O(U (k))\ O(U (A))

for some f in σ1 . Assume again that k is a local field, as before. Then this statement
has a local analogue. The local analogue of the global theta lift of σ1 to Sp(n, A)
is the unique smooth representation Θ(σ1 ) of Sp(n, k) such thatωn (σ1 ) ∼= Θ(σ1 ) ⊗ σ1
as Sp(n, k) × O(X) representations; here, ωn (σ1 ) is the quotient of ωn by the intersection of all the kernels of the elements of HomO(X) (ωn , σ1 ). See [R] for a
discussion. In particular, Θ(σ1 ) is nonzero if and only if σ1 is in Rn (O(X)), and if
σ1 is in Rn (O(X)), then Θ(σ1 ) has a unique nonzero irreducible quotient, which isθ(σ1 ).
Lemma 6.1. Let σ1 be in Irr(O(X)), and let u in X n be nondegenerate. Let
β = (ui , uj ). Then
HomNn (Θ(σ1 ), ψβ ) ∼= HomO(U ) (σ1∨ , 1)
as C vector spaces.

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)799

Proof. By Frobenius reciprocity, as in 2.29 of [BZ],∼ HomO(U ) (1, (σ ∨ )|∨HomO(U) (σ ∨ , 1) =1

1

O(U ) )O(X)∼
= HomO(X) (c-IndO(U ) 1, σ1 )∼ HomO(X) (S(O(X) · u), σ1 ).
=Here, the last statement follows by 1.6 of [BZ]. By Lemma 2.3 of [KR], there is an
O(X) isomorphism(ωn )N ,ψ ∼= S(O(X) · u),nβwhere (ωn )Nn ,ψβ is the Jacquet module of ωn with respect to Nn and ψβ . We thus
obtainHomO(U) (σ ∨ , 1) ∼= HomO(X) ((ωn )N ,ψ , σ1 )
1

n

β

∼
= HomNn ×O(X) ((ωn )Nn ,ψβ , ψβ ⊗ σ1 )∼= HomN ×O(X) (ωn , ψβ ⊗ σ1 )n∼
= HomNn ×O(X) (ωn (σ1 ), ψβ ⊗ σ1 )∼ HomN ×O(X) (Θ(σ1 ) ⊗ σ1 , ψβ ⊗ σ1 )=n∼
= HomNn (Θ(σ1 ), ψβ ).

Corollary 6.2. Suppose that σ is in Irr(GO(X)) and u in X n is nondegenerate.
If
HomO(U ) (σ ∨ , 1) 6= 0,
then σ is in Rn (GO(X)).
Proof. Let σ1 in Irr(O(X)) be an irreducible constituent of σ|O(X) such that
HomO(U ) (σ1∨ , 1) 6= 0.
By Lemma 6.1, we have in particular, Θ(σ1 ) 6= 0. This implies that σ1 is in
Rn (O(X)), and hence σ is in Rn (GO(X)).
Lemma 6.3. Suppose that π is in Irr(GSO(X)), and u in X n is nondegenerate.
Assume that n = 1, 2 or 3. ThenHomSO(U ) (π ∨ , 1) 6= 0 =⇒ HomSO(X) (ωn , π) 6= 0.
Proof. Let β = (ui , uj ). Since n = 1, 2 or 3, it follows that SO(X) · u = O(X) · u.
Hence by an argument as in the proof of Lemma 6.1,
HomSO(U ) (π ∨ , 1) ∼= HomSO(X) (S(O(X) · u), π)∼= HomSO(X) ((ωn )Nn ,ψβ , π)∼
= HomN ×SO(X) (ωn , ψβ ⊗ π).
nThis implies the lemma.
Lemma 6.4. Let π be in Irr(GSO(X)). If π is generically distinguished, then π +is in R1 (GO(X)).
Proof. By Proposition 4.1 and Theorem 5.3, π ∨ is also generically distinguished.
Hence, HomSO(Y ) (π ∨ , 1) 6= 0. By Lemma 6.3 we have HomSO(X) (ωn , π) 6= 0. By
Theorem 4.4, π + is in R1 (GO(X)).

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Lemma 6.5. Suppose that d 6= 1. Let π in Irr(GSO(X)) be one dimensional. If π
is regular, then π + is in R2 (GO(X)). If π is invariant, then π + is in R1 (GO(X))
and π − is not in R1 (GO(X)) and R2 (GO(X)).
Proof. Suppose first that π = π(β, χ) is regular. By Theorem 4.4, it suffices to showthat there is a nonzero SO(X) map from ω2 to π. To prove this we show first that
there exist generically distinguished elements π1 and π2 of Irr(GSO(X)) such that
HomGSO(X) (π1 ⊗C π2 , π) 6= 0. By Proposition 3.2, there exists a nontrivial quadratic
character µ different from ωK/k such that χ = β|k× µ. Since β 2 = χ ◦ NK
k , we haveKβ(β ◦ −)−1 = µ ◦ NK
.
Let
π
=
π(π(1,
µ
◦
N
),
µ)
and
π
=
π(π(β,β◦−), β|k× ).
1
2
kk
Then π1 and π2 are generically distinguished. We have π2 ⊗C π −1 = π1∨ , which
proves our claim.
By Lemma 6.4, π1+ and π2+ are contained in R1 (GO(X)). It follows that each
irreducible constituent of π1 |SO(X) and π2 |SO(X) is a nonzero quotient of ω1 |SO(X) .
Since ω2 ∼= ω1 ⊗C ω1 as representations of O(X), by tensoring and composing we
obtain a nonzero SO(X) map from ω2 to π.
Now suppose that π is invariant. The proof that σ = π + is in R1 (GO(X)) has
several steps. We first claim that 1 is contained in RX (Sl(2, k)). To see this, notethat π(ωK/k , 1)|Sl(2,k) has an irreducible component π 0 such that π 0 is in RH (Sl(2, k))and π 0∨ is in R(K,NK(Sl(2, k)); here, π(ωK/k , 1) is the irreducible principal seriesk )representation of Gl(2, k). See, for example, our summary in section 7 of the theta
correspondence for Sl(2, k) and O(V ), when V is two dimensional. Since X = H ⊥∼as representations of Sl(2, k). Hence,(K, NKk ), it follows that ω1 = ωH ⊗C ω(K,NKk )there is a nonzero Sl(2, k) map of ω1 onto 1.
Using that 1 is in RX (Sl(2, k)), we will show that π(1, ωK/k )+ is in R1 (GO(X)).
It is a straightforward exercise using Corollary 2.6 of [K] to verify that θ(1) in
1/2−1/2
Irr(O(X)) is an irreducible subquotient of π(%(| |K , | |K ), ωK/k )0 |O(X) , where
1/2
−1/2
1/2
−1/2
π(%(| |K , | |K ), ωK/k )0 is an extension of π(%(| |K , | |K ), ωK/k ) to GO(X); here
1/2

−1/2

1/2

−1/2

ρ(| |K , | |K ) is the representation of Gl(2, K) induced from | |K and | |K .
The notation is as in [GL]. Hence, θ(1) is either an irreducible constituent of
1/2−1/2
π(σ(| |K , | |K ), ωK/k )± |O(X) or θ(1) = π(1, ωK/k )± |O(X) . The first possibility
contradicts either our summary in section 7 of M. Cognet’s results [Co] or Theorem
4.4. It follows that θ(1) = π(1, ωK/k )± |O(X) . Assume θ(1) = π(1, ωK/k )− |O(X) .
Using that there is a nonzero O(X) map from ω1 to π(π(1, 1), ωK/k )+ |O(X) from
Lemma 6.4 we obtain a nonzero O(X) map from ω2 ∼= ω1 ⊗C ω1 toπ(π(1, 1), ωK/k )+ ⊗C π(1, ωK/k )− .
Making use of the explicit construction of the nonzero SO(Y ) invariant functionalon π(π(1, 1), ωK/k ), one can verify thatπ(π(1, 1), ωK/k )+ ⊗C π(1, ωK/k )− = π(π(1, 1), 1)− ,so that π(π(1, 1), 1)− is in R2 (GO(X)). This contradicts Theorem 4.3. Hence,θ(1) = π(1, ωK/k )+ |O(X) , and π(1, ωK/k )+ is in R1 (GO(X)).
To see now that for an arbitrary one dimensional, invariant, but not genericallyK+
distinguished π(α ◦ NK
k , χ), we have π(α ◦ Nk χ) in R1 (GO(X)), note thatπ(α ◦ NkK , χ)+ = π(1, ωK/k )+ ⊗C (α ◦ λ).

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)801

Here, λ is the similitude factor map from Section 1. Since
HomR (ω1 , θ(π(1, ωK/k )+ ) ⊗C π(1, ωK/k )+ ) 6= 0,
it follows that+
HomR (ω1 , (θ(π(1, ωK/k )+ ) ⊗C (α−1 ◦ λ)) ⊗C π(α ◦ NK
k , χ) ) 6= 0.
+
Hence, π(α ◦ NK
k , χ) is in R1 (GO(X)).
Finally, we show that π − is not in Rn (GO(X)) for n = 1 and 2. Suppose that π −−1 +is in Rn (GO(X)) for n = 1 or 2. From above, π(α−1 ◦ NK
) is in R1 (GO(X)).
k ,χ
Since ωn+1 = ωn ⊗C ω1 as representations of O(X),
−
−1
−1 +
, ◦ NK) = signπ(α ◦ NKk , χ) ⊗C π(αk ,χis in Rn+1 (GO(X)). This contradicts the appendix of [Ra], since n + 1 < 4.
Lemma 6.6. Let π ∈ Irr(GSO(X)). In the case d 6= 1 assume that π is infinite
dimensional. Then there exists a nondegenerate z in X 2 with stabilizer SO(Z) in
SO(X) such that
HomSO(Z) (π, 1) 6= 0.
Proof. Suppose first d = 1 and  = (1). Let π = π(τ, τ 0 ). Suppose that τ and τ 0
are infinite dimensional. Let

 

0 1
0 0
z=
⊕
.
0 0
1 0
Then z is nondegenerate, and the stabilizer of z is  −10a 0a) : a ∈ k × }.
SO(Z) = {ρ(
,
0
1
0 1
We will use the Kirillov models K(τ, ψ) and K(τ 0 , ψ) of τ and τ 0 with respect toour additive character ψ, respectively. Let n be so large thatZf (x)|x|n dx
k×converges absolutely for f in K(τ, ψ) and f in K(τ 0 , ψ). Define L : π → 1 byZ
Z
L(f ⊗ f 0 ) =f (x)|x|n dx ·f 0 (x)|x|n dx.
k×k×Then L is a well defined nonzero C linear map, and L is SO(Z) invariant.
Suppose next that exactly one of τ and τ 0 , say τ , is infinite dimensional. Since0τ is finite dimensional, τ 0 is one dimensional, and there exists a quasi-character β 0
of k × such that τ 0 = β 0 ◦ det. By hypothesis, β 02 = ωτ 0 = ωτ . Suppose that τ is asupercuspidal or special representation. We claim thatZ
β 0 (x)−1 f (x) dx
k×converges absolutely for f in K(τ, ψ). This is clear if τ is supercuspidal. If τ is the
special representation σ(µ1 , µ2 ) with µ1 = µ2 | |, then this follows from the estimate|πk |1/2 |β 0 (πk )|−1 |µ1 (πk )| = |πk | < 1. Now define L : π → 1 byZ
L(f ⊗ z) = z
β 0 (x)−1 f (x) dx.
k×

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BROOKS ROBERTS

Then L is a nonzero element of HomSO(Z) (π, 1). Suppose that τ is a principal
series representation. In this case, we require another nondegenerate element of
X 2 . Every quadratic extension E of k is contained in M2 (k) as a k algebra, and forevery quadratic extension E of k contained in X = M2 (k), Gal(E/k) = {1, ∗}, andthere exists a nondegenerate z in X 2 such that SO(Z) = {ρ(x, x∗−1 ) : x ∈ E × }.
Fix a quadratic extension E of k in X and such a z in X 2 . Let α be the quasicharacter of E × defined by α(x) = β 0 (det(x)). Then α extends ωτ . By [T], we have
HomE × (τ, α) 6= 0 if and only if (BCE/k (τ ) ⊗C α−1 , 1/2, ψE ) = ωτ (−1). By Lemma
14 of [HST], (BCE/k (τ )⊗C α−1 , 1/2, ψE ) = ωτ (−1), so that HomE × (τ, β 0 ◦det) 6= 0.
Let f in HomE × (τ, β 0 ◦ det) be nonzero. Define L : π → 1 by L(v ⊗ z) = zf (v).
Then L is a nonzero element of HomSO(Z) (π, 1).
Suppose that τ and τ 0 are both finite dimensional, i.e., one dimensional. Let β
and β 0 be quasi-characters of k × such that τ = β ◦ det and τ 0 = β 0 ◦ det. Sinceωτ = ωτ 0 , we have β 2 = β 02 . This implies that β = β 0 or β = ωE/k β 0 for somequadratic extension E of k, since the residual characteristic of k is odd. Let E be×contained in X and let z in X 2 be as above. Since det(x) = NE
k (x) for x in E , itfollows that HomSO(Z) (π, 1) 6= 0.
Now suppose d = 1 and  = −(1). Since SO(X) is compact, it will suffice
to show that there exists nonzero v in π and nondegenerate z in X 2 such that
π(h)v = v for h in SO(Z). Since for every quadratic extension E of k we haveagain that E is contained in D as a k algebra, Gal(E/k) = {1, ∗}, and there exists
a nondegenerate z in X 2 such that SO(Z) = {ρ(x, x∗−1 ) : x ∈ E × }, to prove theexistence of the required v and z it will suffice to show that there exists a quadratic
extension E of k contained in D, a quasi-character φ of E × , and nonzero vectors
w in τ and w0 in τ 0 such that τ (x)w = φ(x)w and τ 0 (x)w0 = φ(x∗ )w0 for x in E × .
If τ and τ 0 are one dimensional, then an argument as in the case  = (1) works.
Suppose dim τ > 1 and dim τ 0 > 1. We will use terminology and results from[T]. We first assert that we can assume that τ and τ 0 are minimal. To see this, let
α = ωτ = ωτ 0 . Consider α|1+πk Ok . For some large n, we can regard α as a characterof 1 + πk Ok /1 + πkn Ok . This is a finite group of odd order. It follows that squaring
is an automorphism of the group of characters of this group. Hence, there exists
a quasi-character η of k × such that η 2 = α on 1 + πk Ok . Consider τ ⊗C η −1 andτ 0 ⊗C η −1 . The common central character of these representations has conductor
less than or equal to 1. Since any element of Irr(D× ) of dimension larger than 1with central character of conductor less than or equal to 1 is minimal, τ ⊗C η −1and τ 0 ⊗C η −1 are minimal. Since our claim holds for τ ⊗C η −1 and τ 0 ⊗C η −1 if andonly if it holds for τ and τ 0 , we may assume that τ and τ 0 are minimal.
Let JL(τ ) and JL(τ 0 ) be the representations corresponding to τ and τ 0 under theJacquet-Langlands correspondence, respectively. Since dim τ > 1 and dim τ 0 > 1,
these representations are supercuspidal. Let a(JL(τ )) and a(JL(τ 0 )) be the conductors of JL(τ ) and JL(τ 0 ), respectively. Without loss of generality, we may assume
that dim(τ ) ≥ dim(τ 0 ). Using the formulas for dim τ and dim τ 0 in terms of a(JL(τ ))and a(JL(τ 0 )), respectively, one can show that a(JL(τ )) ≥ a(JL(τ 0 )). Note that theformula in [T] for dim τ when a(JL(τ )) is odd appears incorrectly: it should be(q + 1)q (c−3)/2 instead of (q + 1)(c−3)/2 . Let E be a quadratic extension of k whose
ramification index e has the same parity as a(JL(τ )). Let S be the set of all quasicharacters of E × whose conductors are less than or equal to e(a(JL(τ )) − 1)/2 andwhich extend α, and let S 0 be the set of all quasi-characters of E × whose conduc-THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)803

tors are less than or equal to [e(a(JL(τ 0 )) − 1)/2 + 1/2] and which extend α. Sincea(JL(τ )) ≥ a(JL(τ 0 )) we have S 0 ⊂ S. By the proof of Lemma 3.2 of [T], τ |E ×
is the direct sum of the elements of S. By the proof of Lemma 3.1 of [T] every
quasi-character of E × that occurs in τ 0 |E × is contained in S 0 . It follows that there
exists a quasi-character φ of E × that occurs in τ |E × and τ 0 |E × . Since the conductorof φ ◦ ∗ is the same as the conductor of φ, it follows that φ ◦ ∗ also occurs in τ |E × ,which proves our claim.
The case when, say, dim(τ ) > 1 and dim(τ 0 ) = 1 remains. Let τ 0 = β 0 ◦ N.
Then β 02 = α. It follows that the common central character of τ ⊗C β 0−1 andτ 0 ⊗C β 0−1 = 1 is trivial. Thus, we may assume that τ is minimal and τ 0 = 1. Let
S be as in the last paragraph. Since α is trivial, it follows that the trivial characterof E × lies in S, and so we can take φ to be the trivial character of E × .
Suppose now d 6= 1. Let π = π(τ, χ). By assumption, τ is infinite dimensional.
Let
√  


0 0
0
d
⊕ √
.
z=
0 0
d 0
A computation shows that
SO(Z) = {ρ(1,


u
0


0
) : u ∈ ker(NKk )}.
1Since SO(Z) is compact it will suffice to show that there exists a nonzero vector v
in τ such that


u 0
τ
v=v
0 1
for u in ker(NKk ). We will use the Kirillov model K(τ, ψK ) of τ . Let f be the charK
acteristic function of O×
K . Then f is in K(τ, ψK ), and since ker(Nk ) is contained
×K
×
in OK , we have f (ux) = f (x) for x in K and u in ker(Nk ). Thus, f is the desired
vector.
Lemma 6.7. Suppose that d 6= 1. Let π be in Irr(GSO(X)). Assume that π
is infinite dimensional, invariant, but not distinguished. Let π1 and π2 be the two
extensions of π to GO(X). Then there exists a nondegenerate z in X 2 with stabilizer
O(Z) in O(X) such that
HomO(Z) (π1 , 1) 6= 0,

HomO(Z) (π2 , 1) 6= 0.

Proof. Let π = π(τ, χ). Then τ is infinite dimensional. Let the notation be as insection 5. Let


 √
d
0
1 0
√
.
z=
⊕
0 1
0 − d
Then


a
SO(Z) = {ρ(a,00) : a ∈ k × },1

and O(Z) is generated by SO(Z) and


0 1
ρ(1,
)h0 .
−1 0
Since π is invariant, by Proposition 3.1, τ is Galois invariant. From the explicit
form of K(τ, ψK ) it follows that K(τ, ψK ) is invariant under composition by −, and804

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a computation shows that (τ (g)f ) ◦ − = τ (g)(f ◦ −). We may assume that π1 (h0 )is given by π1 (h0 )f = f ◦ − and π2 (h0 ) is given by π2 (h0 )f = −f ◦ −. Since π is
not distinguished, by Proposition 4.1 we have that HomGl(2,k) (τ, χ ◦ det) = 0. By
Theorem 5.3, it follows that τ is not the base change of an element of Irr(Gl(2, k))with central character χωK/k . In particular, τ is not π(µ1 , µ2 ) for some Galois
invariant quasi-characters µ1 and µ2 of K × . Let ζ be a quasi-character of K × that
extends χ. By Lemma 5.1,
ZZ(g, ζ −1 , f, 1/2) =τ (g)f (x)ζ(x)−1 dx
K×converges absolutely for all g in Gl(2, K) and f in K(τ, ψK ). Define Lζ : π → 1 byLζ (f ) = Z(1, ζ −1 , f, 1/2).
Then Lζ is nonzero, and a computation shows that Lζ is in HomSO(Z) (π, 1). Moreover,




0 1
0 1
Lζ (π1 (ρ(1,)h0 )f ) = Z(, ζωτ−1 , f, 1/2),
−1 0
−1 0
and






0 1
0 1
)h0 )f ) = −Z(, ζωτ−1 , f, 1/2),Lζ (π2 (ρ(1,
−1 0
−1 0

for f in π. By the local functional equation for τ and Lemma 5.2, we thus have
0 1Lζ (π1 (ρ(1,)h0 )f ) = (τ ⊗C ζ −1 , 1/2, ψ)Lζ (f )
−1 0
and




0 1
Lζ (π2 (ρ(1,)h0 )f ) = −(τ ⊗C ζ −1 , 1/2, ψ)Lζ (f )−1 0

for f in K(τ, χ). Since τ is not the base change of an element of Irr(Gl(2, k)) withcentral character χωK/k , by Lemma 14 of [HST], there exist quasi-characters ζ andζ 0 of K × extending χ such that(τ ⊗C ζ −1 , 1/2, ψ) = χ(−1),

(τ ⊗C ζ 0−1 , 1/2, ψ) = −χ(−1).

This completes the proof.
Proof of Theorem 6.8. The only if part of the theorem follows from Theorem 4.3
and Lemma 6.5.
To prove the if part of the theorem, let π be in Irr(GSO(X)). We need to showthat if π is regular or π is invariant and distinguished, then π + is in R2 (GO(X)),
and if π is invariant but not distinguished, then both extensions of π to GO(X) lie
in R2 (GO(X)).
Suppose d = 1. Then this follows from Lemma 6.6, Lemma 6.3 and Theorem
4.4.
Suppose now d 6= 1. Suppose first π is regular or invariant and distinguished. If
π is one dimensional, then we use Lemma 6.5. If π is infinite dimensional, then anargument as in the case d = 1 is sufficient. Finally, suppose π is invariant but notdistinguished. By definition, π is infinite dimensional. By Lemma 6.7 and Corollary
6.2, the extensions π1 and π2 of π to GO(X) are in R2 (GO(X)).

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)805

Proof of Corollary 6.9. Let σ1 be in Irr(O(X)). Suppose that σ1 is in R2 (O(X)).
Suppose that σ1 is an irreducible constituent of π − |O(X) for some distinguished π in
Irr(GSO(X)). Then by definition, π − is in R2 (GO(X)), contradicting Theorem 6.8.
Suppose that σ1 is not an irreducible constituent of π − |O(X) for all distinguished
π in Irr(GSO(X)). By [GK], there exists a σ in Irr(GO(X)) such that σ|O(X) has
σ1 as an irreducible constituent. By Theorem 6.8, σ is in R2 (GO(X)). By Lemma
4.2 of [R], σ1 is in R2 (O(X)).
7. A case of a conjecture of Kudla
S.S. Kudla has made some important conjectures about the first occurrence ofa representation in the theta correspondence. In the introduction to the paper we
described one of S.S. Kudla’s conjectures. This conjecture is known to be true whendimk X = 0 and 2. We will prove this conjecture when dimk X = 4. We will also
show that the isotropy assumption in the definition of a generically distinguishedrepresentation in the case d 6= 1 is unnecessary. This will be a consequence of anew proof of a result of J. Hakim and D. Prasad on distinguished representations.
Tables summarizing the main results of this paper appear at the end of this section.
There is also a conjecture of S.S. Kudla for elements of Irr(Sp(n, k)) which wewill give for completeness; we will partially prove and use a case of the conjecture.
To state this conjecture we need some more notation. Fix d in k × /k ×2 . Then there
are, up to equivalence, exactly two anisotropic even dimensional symmetric bilinear
spaces X+ and X− of discriminant d. From X+ and X− we can create two seriesof even dimensional symmetric bilinear spaces by adding hyperbolic planes to X+
and X− . For π in Irr(Sp(n, k)), let m+ (π) be the smallest nonnegative even integerm such that π occurs in the theta correspondence with the m dimensional spacewith anisotropic component X+ ; define m− (π) similarly.
Conjecture 7.2 (S.S. Kudla). If π is in Irr(Sp(n, k)), thenm+ (π) + m− (π) = 4n + 4.
Recently, S.S. Kudla and S. Rallis have announced a proof of the equality ofthis conjecture for a set of representations π in Irr(Sp(n, k)) that includes all thesupercuspidal representations. One can make completely analogous definitions andconjectures for the theta correspondence for similitudes.
The following lemma will used in the proof of Theorem 7.4.
Lemma 7.3. Let π be in Irr(Sp(1, k)) = Irr(Sl(2, k)). Thenm+ (π) + m− (π) ≥ 8.
Proof. Suppose that m+ (π)+m− (π) < 8 for some π in Irr(Sl(2, k)). We will obtain acontradiction. Let X1 and X2 be the symmetric bilinear spaces of dimension m+ (π)
and m− (π) with anisotropic components X+ and X− , respectively. By hypothesis,
we have surjective Sl(2, k) maps of ωX1 and ωX2 onto π. Let g0 in Gl(2, k) be suchthat det(g0 ) = −1. By the theorem on page 91 of [MVW], g0 · π ∼= π ∨ . It follows∨
that there is a surjective Sl(2, k) map of g0 · ωX2 onto π . Now g0 · ωX2 ∼= ω−X2
as representations of Sl(2, k), where −X2 is the symmetric bilinear space with thesame underlying vector space as X2 , but with bilinear form multiplied by −1. SinceωX1 ⊗C ω−X2 ∼= ωX1 ⊥−X2 as representations of Sl(2, k), and since there is a nonzero
Sl(2, k) map from π ⊗C π ∨ to 1, it follows that
HomSl(2,k) (ωX1 ⊥−X2 , 1) 6= 0.

806

BROOKS ROBERTS

Now the anisotropic component of X1 ⊥ −X2 is the four dimensional anisotropic
symmetric bilinear space Xa = X(1, −(1)). Since the dimension of X1 ⊥ −X2 isless than 8, X1 ⊥ −X2 is Xa or Xa ⊥ H. Suppose X1 ⊥ −X2 ∼= Xa . Then 1 is in
RXa (Sl(2, k)). This is a contradiction, since the restriction of ωXa to Sl(2, k) is the
direct sum of square-integrable representations of Sl(2, k), and hence RXa (Sl(2, k))contains only square-integrable representations. See the discussion in [G], for example. Suppose X1 ⊥ −X2 ∼= Xa ⊥ H. Then ωX1 ⊥−X2 ∼= ωXa ⊗C ωH as representations of Sl(2, k). By the just mentioned fact concerning the decomposition of
the restriction of ωXa to Sl(2, k), there exists a square-integrable representation τ
in Irr(Sl(2, k)) such that
HomSl(2,k) (τ ⊗C ωH , 1) 6= 0.
This implies that there is a nonzero Sl(2, k) map from ωH to τ ∨ , so that τ ∨ isin RH (Sl(2, k)). This is a contradiction, since RH (Sl(2, k)) contains no squareintegrable representations. See, for example, the summary following the lemma.
Suppose X is again as in defined in Section 2. To prove Conjecture 7.1 inthis case, we need to understand R1 (GO(X)) and R3 (GO(X)). To characterize
R1 (GO(X)) we need to recall some facts about the theta correspondence when thedimension of the underlying bilinear spaces is two and about the theta correspondence between Irr(GO(X)) and Irr(Gl(2, k)+ ) in the case d 6= 1.
Let V be a nondegenerate two dimensional symmetric bilinear space of discriminant d. Then GSO(V ) is abelian, and all the elements of Irr(GSO(V )) are one
dimensional. We define regular and invariant representations exactly as in section
3. If α in Irr(GSO(V )) is regular, α+ will again denote the induced representationof α to GO(V ). Moreover, we say that α in Irr(GSO(V )) and is distinguished, if
and only if
HomSO(V ) (α, 1) 6= 0.
Thus, SO(V ) plays the role that SO(Y ) did in section 4, and if α is in Irr(GSO(V ))and is distinguished, then we define α+ and α− just as in section 4. A result entirely
analogous to the main theorem holds: If β is in Irr(GO(V )), then β is in R1 (GO(V ))if and only if β is not of the form α− for some distinguished α in Irr(GSO(V )).
Moreover, by Theorem 1.9 of [Ca], Conjecture 1.3 (theta dichotomy) holds for
X = V and 2n = 2, and the remarksp
preceding Conjecture 1.3 apply. If one makes
the identification of V with K = k( disc(V )), then elements of GSO(V ) can be
identified with quasi-characters of K × . The map that takes a quasi-character α of
K × to θ(α+ )∨ is just the usual map that associates to a quasi-character an elementof Irr(Gl(2, k)). In particular, if V is anisotropic, and α = χ ◦ NK
k is invariant,then θ(α+ )∨ = π(χ, χωK/k ); and if V is isotropic, so that K × = k × × k × andα = (α1 , α2 ), then θ(α+ )∨ = π(α1 , α2 ). See, for example, [G].
The case when V is anisotropic contains information about the restriction ofrepresentations of Gl(2, k) which we will use in the proof of the next theorem.
Let π be in Irr(Gl(2, k)). It is well known that the restriction of π to Sl(2, k) is
multiplicity free and that π|Sl(2,k) is reducible if and only if π is a theta lift ofan element of Irr(GO(V )) for some anisotropic V . See [Sh]. Let π be a theta
lift of σ in Irr(GO(V )) with V anisotropic. Then from Lemma 4.2 of [R] and theremarks in section 1 it follows that the restriction of π to Sl(2, k) has two irreducible
components if and only if σ  α+ with α such that α|SO(V ) 6= 1, and α|2SO(V ) = 1.

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)807

Let α in Irr(GSO(V )) be such that α|SO(V ) 6= 1 and α|2SO(V ) = 1, and assume
π = θ(α+ ). Then again from Lemma 4.2 of [R] the restriction of π to Sl(2, k) has
four components. Finally, from Theorem 1.9 (d) of [Ca] it follows that every such
π, that is, every π in Irr(Gl(2, k)) whose restriction to Sl(2, k) has four components,
is a theta lift from every anisotropic V .
In the case d 6= 1 we also need to make some remarks about the theta correspondence between Irr(GO(X)) and Irr(Gl(2, k)+ ). This was considered in [Co] using
the induced Weil representation Ω of Gl(2, k) × GO(X); see [R] for the definition.
By an argument as in the proof of Proposition 3.5 of [R], as representations ofGl(2, k) × GSO(X),Ω∼= c-IndR0

Gl(2,k)×GSO(X)ω,

where R0 is as in the remark after Theorem 4.4. Using Frobenius reciprocity, themain result of [Co] now states that for every infinite dimensional % in Irr(Gl(2, k)),if BC(%∨ ) is the base change of %∨ to Gl(2, K) and π = π(BC(%∨ ), ωK/k ω%∨ ), then
HomR0 (ω, % ⊗C π) 6= 0.
∨

If BC(% ) is infinite dimensional, then by Proposition 4.1 and Theorem 5.3, π is
generically distinguished. If BC(%∨ ) is one dimensional, then one can check directlythat π is generically distinguished. By the remark following Theorem 4.4, it follows
that
HomR (ω, % ⊗C π + ) 6= 0.
In the case d = 1, the next theorem essentially follows from Lemma 6.1, as one
would expect. In the case d 6= 1, our definition of distinguished representations
is more restrictive, and we need to use explicit knowledge of the relevant theta
correspondence. The extra effort yields Corollaries 7.5 and 7.6.
Theorem 7.4. Let σ be in Irr(GO(X)). Then σ is in R1 (GO(X)) if and only if σ
is of the form π + for some distinguished π in Irr(GSO(X)).
Proof. Suppose that σ is in R1 (GO(X)). Suppose first d = 1. Consider θ(σ) in
Irr(Gl(2, k)). Assume that θ(σ) is infinite dimensional. It is well known that θ(σ)
admits a Whittaker functional, i.e., HomN1 (θ(σ), ψ1 ) 6= 0. By Lemma 4.2 of [R], it
follows that HomN1 (θ(σ1 ), ψ1 ) 6= 0 for some σ1 in Irr(O(X)) which is a constituentof σ|O(X) . Since θ(σ1 ) is a quotient of Θ(σ1 ), it follows that HomN1 (Θ(σ1 ), ψ1 ) 6= 0.
By Lemma 6.1, this implies that HomO(Y ) (σ1∨ , 1) 6= 0 and so HomO(Y ) (σ ∨ , 1) 6= 0.
It is not hard to see that this implies that σ|GSO(X) is irreducible and genericallydistinguished. By Theorem 4.3, σ = π + . Assume that θ(σ) is one dimensional. By[S],  = −(1). By (v) of Theorem 2.2 of [KR] , Θ(1O(X) ) has 1Sl(2,k) as a quotient,i.e., θ(1O(X) ) = 1Sl(2,k) = θ(σ)|Sl(2,k) , so that by Lemma 4.2 of [R] σ|O(X) has 1 asconstituent. This implies that σ = π + for some distinguished π.
Now suppose d 6= 1 and  = (d). In this case, we cannot proceed as in the lastparagraph and use Lemma 6.1. This is because θ(σ) is now in Irr(Gl(2, k)+ ), andwe cannot conclude that HomN1 (θ(σ), ψ1 ) 6= 0. Instead, we must use the explicit
knowledge developed in the paragraphs before the theorem. Suppose first θ(σ)
extends to a representation % of Gl(2, k). Assume that θ(σ) is one dimensional.
×
Then θ(σ) = η ◦ det for some quasi-character η of NKk (K ). Hence, θ(σ)|Sl(2,k) = 1.
We saw in the proof of Lemma 6.2 that θ(1) = π(1, ωK/k )+ |O(X) . By Lemma 4.2of [R], θ(θ(σ))|O(X) = σ|O(X) has π(1, ωK/k )+ |O(X) as an irreducible component.

808

BROOKS ROBERTS+
This implies that σ = π(α ◦ NK
for some one dimensional π(α ◦ NK
k , χ)k , χ) in
Irr(GO(X)). Next, assume that θ(σ) is infinite dimensional. Then if the notation
is as in the discussion preceding the lemma, we find that σ = θ(%|Gl(2,k)+ ) = π + .
Suppose that θ(σ) induces irreducibly to Gl(2, k), and assume that σ is not of
the form π + for some distinguished π. Let % be the induction of θ(σ) to Gl(2, k).
Then % is infinite dimensional. Again, there is a nonzero R0 map from ω to % ⊗C π.
Let g in Gl(2, k) be a representative for the nontrivial coset of Gl(2, k)/ Gl(2, k)+ .
It follows that at least one of

HomR0 (ω, θ(σ) ⊗C π),

HomR0 (ω, gθ(σ) ⊗C π)

is nonzero. If the first space is nonzero, then we find as in the last paragraph thatσ = π + , a contradiction. It follows that the first space is zero and the second isnonzero. This implies that
HomSl(2,k) (ω 0 , θ(σ)) 6= 0,
where ω 0 = g −1 ω is the extended Weil representation corresponding to the otherfour dimensional symmetric bilinear space of discriminant d. Hence, m+ (θ(σ)) ≤ 4
and m− (θ(σ)) ≤ 4. By Lemma 7.3, this implies that m+ (θ(σ)) = m− (θ(σ)) = 4.
It follows that % is not a lift from a two dimensional symmetric bilinear space withdiscriminant d. However, the restriction of % to Gl(2, k)+ is reducible, and so by
our above discussion % is a lift from an anisotropic two dimensional symmetricbilinear space of discriminant different from d. This, along with the fact that % has
a reducible restriction to Gl(2, k)+ , implies that the restriction to Sl(2, k) of % has
four distinct irreducible components. But then by our above remarks, % is a liftfrom a two dimensional symmetric bilinear space of the same discriminant as X.
This is a contradiction.
Corollary 7.5. Assume d 6= 1. Let π in Irr(GSO(X)) be infinite dimensional andinvariant. Let y 0 be anisotropic, and assume that the orthogonal complement Y 0 of
y 0 is anisotropic. If
HomSO(Y 0 ) (π, 1) 6= 0,
then π is generically distinguished.
Proof. First, it is not difficult to see that we may assume that y 0 = S −1 (1), where S
is the intertwining similitude from section 2. It follows that SO(Y 0 ) is the group ofρ(det(g), g) for g in D× , with D× embedded in Gl(2, K) as in section 2. Moreover,O(Y 0 ) is generated by SO(Y 0 ), and h00 = S −1 ◦ ∗ ◦ S. Second, we note that Proposition 4.1 and Lemma 4.2 hold for Y 0 in place of Y . Since Proposition 4.1 holds,
we can define π +0 as we defined π + in section 4, with Y 0 in place of Y . An analysisof the proofs of Theorem 4.4 and Lemma 6.1 shows that these results still hold, sothat π +0 is in R1 (GO(X)). By Theorem 7.4, π +0 = π 00+ for some distinguished π 00
in Irr(GSO(X)). It follows that π = π 00 , and π is generically distinguished.
When ωτ = 1 and χ = 1, the following result is also a consequence of some of
the results of [H] and [P2]. There, different means are employed.
Corollary 7.6. Let τ in Irr(Gl(2, K)) be infinite dimensional and Galois invariant.
Let ωτ = χ ◦ NK
k . Then
HomD× (τ, χ ◦ det) 6= 0 =⇒ HomGl(2,k) (τ, χ ◦ det) 6= 0.

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)809

Proof. Assume the first homomorphism space is nonzero. ThenHomSO(Y 0 ) (π(τ, χ), 1) 6= 0,0where Y is as in the proof of Corollary 7.5. By Corollary 7.5, π(τ, χ) is genericallydistinguished. The result now follows by Proposition 4.1.
Lemma 7.7. Let σ be in Irr(GO(X)). If σ|O(X) 6= sign, then σ is in R3 (GO(X)).
Proof. By Theorem 6.8 and the principle of persistence [MVW], p. 67, it suffices
to show that if π in Irr(GSO(X)) is distinguished and π|SO(X) 6= 1, then π − is in
R3 (GO(X)). We will use the technique of section 6. Let u1 in X 3 be such that thecomponents of u1 form a basis for the orthogonal complement of k · 1. If d 6= 1,
also let
√ 
 

 √
0
d
0
d
1 0
√ ⊕ √
u1 =
⊕
0 1
d 0
0 − d
and
u2 =

√ 
 

 √
0
d
0
d
1 0
√
⊕ −1 √
.
⊕
0 1
d 0
0 − d
δ

×
Here, δ is a representative for the nontrivial coset of k × / NKk (K ). Then O(Ui ) ={1, hi } is the stabilizer of ui in O(X) where h1 = −h0 and




0 1
0 δ
)h0 ,
h3 = ρ(δ,)h0 .
h2 = ρ(1,
−1 0
−1 0

By Corollary 6.2, it suffices to show that for i = 1, 2 or 3, there exists a nonzero
vector v in the space of π such that π − (hi )v = v; to prove this it suffices to showthat for either i = 1, 2 or 3, π − (hi ) 6= −1. To this end, suppose that π − (hi ) = −1
for i = 1, 2 and 3.
Assume first d = 1. Let π = π(τ, τ ). Since π − (h1 ) = −1, we have π(h) =00π(h0 hh−10 ) for h in GSO(X), so that τ (g) ⊗ τ (g ) = τ (g ) ⊗ τ (g) for g in Gl(2, k)
×or D . This is a contradiction, since by the assumption that π|SO(X) 6= 1 thedimension of σ is larger than one.
Assume next d 6= 1. If π is infinite dimensional, then there is a contradiction as in
the case d = 1. Suppose finally that π = π(α◦NKk , χ) is one dimensional. Using thatπ|SO(X) 6= 1 a computation shows that π − (h2 ) = −π − (h3 ), a contradiction.
Proof of Theorem 7.8. Let π in Irr(GSO(X)) be a constituent of the restriction of
σ to GSO(X). Suppose first π is regular so that σ = π + . By Theorem 7.4, we haven(σ) = n(σ ⊗C sign) ≥ 2. By Theorem 6.8, it follows that n(π + ) = n(π + ⊗C sign) =2.
Suppose next that π is distinguished. Without loss of generality, we may assume
that σ = π + . Suppose π|SO(X) 6= 1. Then by Theorem 7.4, n(σ) = 1 and by
Theorem 4.3, Lemma 6.5 and Lemma 7.7, n(σ ⊗C sign) = n(π − ) = 3. Supposethat π|SO(X) = 1. Then σ|O(X) = 1, and by the appendix of [Ra], n(σ) = 0 andn(σ ⊗C sign) = n(π − ) = 4.
Finally, suppose that d 6= 1 and π is invariant but not distinguished. Then by
Theorem 7.4 and Theorem 6.8, n(σ) = n(σ ⊗C sign) = 2.
Proof of Corollary 7.9. Let σ1 be in Irr(O(X)). As we pointed out in the proof of
Theorem 7.8, there exists a σ in Irr(GO(X)) such that σ1 is a constituent of σ|O(X) .

810

BROOKS ROBERTS

By definition and Lemma 4.2 of [R], n(σ1 ) = n(σ). The corollary follows now from
Theorem 7.8.
The following table summarizes the results when d = 1.
d = 1, σ ∈ Irr(GO(X))σn(σ)σ|O(X) = 1
0+
σ|O(X) 6= 1, σ = π , π invariant
1
σ = π + , π regular2−
σ|O(X) 6= sign, σ = π , π invariant3
σ|O(X) = sign
4n(σ ⊗C sign)
4
3
2
1
0

The next table summarizes the information when d 6= 1.
d 6= 1, σ ∈ Irr(GO(X))σn(σ)σ|O(X) = 1
0
σ|O(X) 6= 1, σ = π + , π distinguished1
σ|GSO(X) invariant, not distinguished2
σ = π + , π regular2−
σ = π , π distinguished3
σ|O(X) = sign
4n(σ ⊗C sign)
4
3
2
2
1
0

References
[BZ]
[C]

[Ca]
[Co]
[F]
[G]
[GK]
[GL]

[GRS]
[H]

I.N. Bernshtein and A.V. Zelevinskii, Representations of the group Gl(n, F ) where F isa nonarchimedean local field, Russian Math. Surveys 31 (1976), 1–68. MR 54:12988
P. Cartier, Representations of p-adic groups: a survey, in Automorphic Forms, Representations, L-functions, Proc. Symposia Pure Math. vol. XXXIII - Part 1, American
Mathematical Society, Providence, 1979. MR 81e:22029
W. Casselman, On the representations of Sl2 (k) related to binary quadratic forms, Amer.
J. Math. 94 (1972), 810–834. MR 47:6948
M. Cognet, Representation de Weil et changement de base quadratique, Bull. Soc. Math.
France 113 (1985), 403–457. MR 88h:22029a
Y. Flicker, On distinguished representations, J. Reine Angew. Math. 418 (1991), 139–172.
MR 92i:22019
S.S. Gelbart, Automorphic Forms on Adele Groups, Annals of Mathematics Studies 83,
Princeton University Press, Princeton, 1975. MR 52:280
S.S. Gelbart and A.W. Knapp, L-indistinguishability and R groups for the special linear
group, Adv. in Math. 43 (1982), 101–121. MR 83j:22009
P. Géradin and J.P. Labesse, The solution of a base change problem for Gl(2) (following
Langlands, Saito, Shintani), in Automorphic Forms, Representations, L-functions, Proc.
Symposia Pure Math. vol. XXXIII - Part 2, American Mathematical Society, Providence,
1979. MR 82e:10047
S. Gelbart, J. Rogawski and D. Soudry, On periods and algebraic cycles for U(3), Israel
J. of Math. 83 (1993), 213–252. MR 95a:11047
J. Hakim, Distinguished p-adic representations, Duke Math. J. 66 (1991), 1–22. MR
92c:22037

THE NONARCHIMEDEAN THETA CORRESPONDENCE FOR GSp(2) AND GO(4)811

M. Harris and S.S. Kudla, The central critical value of a triple product L-function, Annalsof Math. 133 (1991), 605–672. MR 93a:11043[HKS] M. Harris, S.S. Kudla and W.J. Sweet, Theta dichotomy for unitary groups, Preprint(1994). MR 96m:11041[HLR] G. Harder, R. Langlands and M. Rapoport, Algebraische Zyklen auf Hilbert-BlumenthalFlachen, J. Reine Angew. Math. 366 (1986), 53–120. MR 87k:11066[HPS] R. Howe and I.I. Piatetski-Shapiro, Some examples of automorphic forms on Sp4 , Duke
Math. J. 50 (1983), 55–106. MR 84m:10019[HST] M. Harris, D. Soudry and R. Taylor, l-adic representations associated to modular forms
over imaginary quadratic fields I: lifting to GSp4 (Q), Invent. Math. 112 (1993), 377–411.
MR 94d:11035[K]
S.S. Kudla, On the local theta correspondence, Invent. Math. 83 (1986), 229–255. MR
87e:22037[KR]
S.S. Kudla and S. Rallis, A regularized Siegel-Weil formula: the first term identity, Annalsof Math. 140 (1994), 1–80. MR 95f:11036[MVW] C. Moeglin, M.-F. Vigneras and J.-L. Waldspurger, Correspondances de Howe sur uncorps p-adique, Lecture Notes in Mathematics 1291, Springer-Verlag, Berlin-HeidelbergNew York, 1987. MR 91f:11040[O]
O.T. O’Meara, Introduction to Quadratic Forms, Springer-Verlag, Berlin-Heidelberg-New
York, 1973. MR 50:269[P1]
D. Prasad, Trilinear forms for representations of Gl(2) and local -factors, Compositio
Math. 75 (1990), 1–46. MR 91i:22023[P2]
D.Prasad, Invariant forms for representations of Gl2 over a local field, Amer. J. Math.
114 (1992), 1317–1363. MR 93m:22011[PSS] I.I. Piatetski-Shapiro and D. Soudry, L and ε factors for GSp(4), J. Fac. Sci. Univ. Tokyo
Sect. IA Math. 28 (1982), 505–530. MR 84m:22029[R]
B. Roberts, The theta correspondence for similitudes, Israel J. of Math. 94 (1996), 285–
317. MR 98a:22007[Ra]
S. Rallis, On the Howe duality conjecture, Compositio Math. 51 (1984), 333–399. MR
85g:22034[Sh]
D. Shelstad, Notes on L-indistinguishability, in Automorphic Forms, Representations,L-functions, Proc. Symposia Pure Math. vol. XXXIII - Part 2, American MathematicalSociety, Providence, 1979. MR 81m:12018[S]
H. Shimizu, Theta series and automorphic forms on Gl(2), J. Math. Soc. Japan 24(1972), 638–683. MR 54:12658[Sc]
W. Scharlau, Quadratic and Hermitian Forms, Springer-Verlag, Berlin-Heidelberg-New
York, 1985. MR 86k:11022[So1]
D. Soudry, The L and γ factors for generic representations of GSp(4, k) × Gl(2, k) overa local nonarchimedean field k, Duke Math. J. 51 (1984), 355–394. MR 86f:22022[So2]
D. Soudry, A uniqueness theorem for representations of GSO(6) and the strong multiplicity one theorem for generic representations of GSp(4), Israel J. of Math. 58 (1987),
257–287. MR 89f:22025[T]
J. Tunnell, Local -factors and characters of Gl(2), Amer. J. Math. 105 (1983), 1277–
1307. MR 86a:22018[V]
M.F. Vignéras, Correspondances entre representations automorphes de Gl(2) sur une
estension quadratic de GSp(4) sur Q, conjecture locale de Langlands pour GSp(4), in The
Selberg Trace Formula and Related Topics, American Mathematical Society, Providence,
R.I., 1986. MR 88e:11118[W]
J.-L. Waldspurger, Demonstration d’une conjecture de duality de Howe dans le case padiques, p 6= 2, in Israel Math. Conf. Proc. vol. 2, 1990, pp. 267–324. MR 93h:22035[HK]

Department of Mathematics, University of Toronto, Toronto, Ontario M5S 3G3,
CanadaCurrent address: Department of Mathematics, Brink Hall, University of Idaho, Moscow, Idaho
83844-1103
E-mail address: brooks@member.ams.org