Exterior Square -Factors for Simple Supercuspidal Representations

Home , James Cogdell

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Conjecture 1.2. Let π1 and π2 be irreducible unitarizable supercuspidal representations of GLn (F ) sharing the same central character. Suppose for any character χ of F ∗, we have γ(s, π χ, ψ)= γ(s, π χ, ψ), 1 × 2 × and γ(s, π , j χ, ψ)= γ(s, π , j χ, ψ), 1 ∧ ⊗ 2 ∧ ⊗ for any 2 j n . Then π = π . ≤ ≤ 2 1 ∼ 2 When j = 2, the twisted exterior square gamma factors of irreducible supercuspidal representations of GLn (F ) exist due to the work of Jacquet-Shalika [JS90] together with Ma- tringe [Mat14] and Cogdell-Matringe [CM15], or the work of Shahidi [Sha90] using the Langlands-Shahidi method. When j = 3, Ginzburg and Rallis [GR00] found an integral representation for the automorphic L-function L(s, π, 3 χ) attached to an irreducible cus- ∧ ⊗ pidal automorphic representation π of GL6 (A) and a character χ of GL1 (A) for some adelic ring A. In general, for j 3, we don’t have an analytic deﬁnition for γ(s, π, j χ, ψ). Therefore, Conjecture 1.2 only≥ makes sense for n =4, 5 and possibly 6 if one can∧ prove⊗ local functional equations for the local integrals coming from [GR00].

Since we have only twisted exterior square gamma factors in general, we want to know which families of supercuspidal representations of GLn (F ) satisfy Conjecture 1.2 when j = 2. We show in the paper that Conjecture 1.2 holds true for simple supercuspidal representations up to a sign as we will explain in the next paragraph. We have already seen that GL1 (F ) twists are enough to distinguish simple supercuspidal representations, see [BH14, Proposition 2.2] and [AL16, Remark 3.18]. Thus, Conjecture 1.2 for simple supercuspidal representations has no context if we still require GL1 (F ) twists. Therefore, we will drop the assumption on the GL1 (F ) twists.

Let o be the ring of integers of F , and p =(̟) is the maximal prime ideal in o generated by a fixed uniformizer ̟. By [KL15], if we fix a tamely ramified central character ω, i.e., ω is trivial on 1 + p, there are exactly n(q 1) isomorphism classes of irreducible simple − supercuspidal representations of GLn (F ), each of which corresponds to a pair (t0,ζ), where 1 t f∗ is a non-zero element in the residue field f of F and ζ is an n-th root of ω(t− ̟), 0 ∈ where t is a lift of t0 to o∗. The main theorem of the paper is the following:

Theorem 1.3. Let π and π′ be irreducible simple supercuspidal representations of GLn (F ) sharing the same central character ω, such that π, π′ are associated with the data (t0,ζ), (t0′ ,ζ′) respectively. Assume that

1) gcd(m 1, q 1)=1 if n =2m, 2) or gcd(−m, q −1)=1 if n =2m +1. −

Suppose for every unitary tamely ramiﬁed character µ of F ∗, we have

2 2 γ(s, π, µ, ψ)= γ(s, π′, µ, ψ). ∧ ⊗ ∧ ⊗ Then t = t′ and ζ = ζ′. Moreover, we have ζ = ζ′ if n =2m +1 is odd. 0 0 ± EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 3

In the case n = 2m, we can only show ζ and ζ′ are equal up to a sign. That is what we mean by saying that Conjecture 1.2 holds true for simple supercuspidal representations up to a sign.

In Section 2, we recall the deﬁnitions of twisted exterior square gamma factors following [JS90, Mat14, CM15]. We then recall some results on simple supercuspidal representations in [KL15]. More importantly from [AL16, Section 3.3], we have explicit Whittaker functions for such simple supercuspidal representations. Using these explicit Whittaker functions, we compute in Section 3 the twisted exterior square gamma factors. Finally in Section 4, we prove our main theorem, Theorem 1.3.

2. Preliminaries and notation

2.1. Notation. Let F be a non-archimedean local ﬁeld. We denote by o its ring of integers, by p the unique prime ideal of o, by f = o/p its residue ﬁeld. Denote q = f . | | Let ν : o f be the quotient map. We continue denoting by ν the maps that ν induces on various groups,→ for example om fm, M (o) M (f), GL (o) GL (f) etc. → m → m m → m Let ̟ be a uniformizer (a generator of p). We denote by , the absolute value on F , normalized such that ̟ = 1 . |·| | | q

Let ψ : F C∗ be a non-trivial additive character with conductor p, i.e. ψ is trivial on p, but not on→o.

2.2. The twisted Jacquet-Shalika integral. In this section, we deﬁne twisted versions of the Jacquet-Shalika integrals, and discuss the functional equations that they satisfy. This will allow us to deﬁne the twisted exterior square gamma factor γ (s, π, 2 µ, ψ) for a ∧ ⊗ generic representation (π,Vπ) of GLn (F ) and a unitary character µ : F ∗ C∗. We will need this for our local converse theorem in Section 4. →

We denote Nm the upper unipotent subgroup of GLm (F ), Am the diagonal subgroup of GL (F ), K = GL (o), the upper triangular matrix subspace of M (F ), and − the m m m Bm m Nm lower triangular nilpotent matrix subspace of M (F ). We have M (F ) = −. m Bm\ m ∼ Nm For the following pairs of groups A B, we normalize the Haar measure on B so that the ≤ compact open subgroup A has measure one: o F , o∗ F ∗, K = GL (o) GL (F ), ≤ ≤ m m ≤ m − (o) − (F ). Nm ≤Nm

Recall the Iwasawa decomposition: GLm (F )= NmAmKm. It follows from this decomposition that for an integrable function f : N GL (F ) C we have m\ m →

1 f (g) d×g = f (ak) δB−m (a) d×ad×k, Nm GLm(F ) Km Am Z \ Z Z 4 RONGQING YE & ELAD ZELINGHER

1 where B GL (F ) is the Borel subgroup, and for a = diag (a ,...,a ), δ− (a) = m ≤ m 1 m Bm aj 1 i

Q Let (π,V π) be an irreducible generic representation of GLn (F ), and denote its Whittaker model with respect to ψ by (π, ψ). Let µ : F ∗ C∗ be a unitary character. We now deﬁne the twisted Jacquet-ShalikaW integrals and their→ duals. These initially should be thought as formal integrals. We discuss their convergence domains later and explain how to interpret them for arbitrary s C. ∈ 1 We have a map (π, ψ) (π, ψ− ), denoted by W W , where π is the contragredient W → W 7→ 1 ι ι t 1 representation, and W is given by W (g)= W (w g ), where w = .. and g = g . n f n . − e 1e Denote by (F m)f the space of Schwartzf functions φ : F m C, that is the space of locally constant functionsS with compact support. →

Suppose n =2m. We deﬁne for s C, W (π, ψ), φ (F m) ∈ ∈ W ∈ S I X g J (s,W,φ,µ,ψ)= W σ m ψ ( trX) 2m I g Nm GLm(F ) m Mm(F ) m − \ B \ Z Z s det g µ (det g) φ (εg) dXd×g, ·| | m where ε = εm = 0 ... 0 1 F , and σ2m is the column permutation matrix corresponding to the permutation ∈ 1 2 ... m m +1 m +2 ... 2m , 1 3 ... 2m 1 | 2 4 ... 2m − | i.e.,

σ2m = e1 e3 ... e2m 1 e2 e4 ... e2m , −

where ei is the i-th standard column vector, for 1 i 2m. In this case, we deﬁne the dual Jacquet-Shalika as ≤ ≤

Im 1 1 J˜(s,W,φ,µ,ψ)= J 1 s, π W, ψφ,µ− , ψ− , − Im F where J on the right hand side is the Jacquet-Shalikae integralf for π, and

m ψφ (y)= q 2 φ (x) ψ ( x, y ) dx e F m h i ZF is the Fourier transform, normalized such that it is self dual (here , is the standard bilinear form on F m). h· ·i

Suppose n =2m + 1. We deﬁne for s C, W (π, ψ), φ (F m) ∈ ∈ W ∈ S EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 5

J (s,W,φ,µ,ψ)= Nm GLm(F ) M1×m(F ) m Mm(F ) Z \ Z ZB \ Im X g Im W σ2m+1 Im g Im ·   1  1  Z 1 s 1 ψ (trX) det g − µ (detg) φ (Z) dXdZd  ×g,  · − | | where σ2m+1 is the column permutation matrix corresponding to the permutation 1 2 ... m m +1 m +2 ... 2m 2m +1 , 1 3 ... 2m 1 | 2 4 ... 2m | 2m +1 − | | i.e.

σ2m+1 = e1 e3 ... e2m 1 e2 e4 ... e2m e2m+1 . − In this case, we deﬁne the dual Jacquet-Shalika as

Im 1 1 J˜(s,W,φ,µ,ψ)= J 1 s, π Im W, ψφ,µ− , ψ− .  −  1 F   e   f  In both cases, we denote J (s,W,φ,ψ)= J (s, W, φ, 1, ψ) and J˜(s,W,φ,ψ)= J˜(s, W, φ, 1, ψ), where 1 denotes the trivial character F ∗ C∗. → We now list properties of the twisted Jacquet-Shalika integrals, some of which are only proven for the (untwisted) Jacquet-Shalika integrals in the literature.

From now and on suppose n =2m or n =2m + 1. Theorem 2.1 ([JS90, Section 7, Proposition 1, Section 9, Proposition 3]). There exists m rπ, 2 R, such that for every s C with Re(s) >rπ, 2 , W (π, ψ) and φ (F ), the integral∧ ∈ J (s,W,φ,µ,ψ) converges∈ absolutely. ∧ ∈ W ∈ S

Similarly, there exists a half left plane Re (s) ∈ W ∈ S s 7→ ∈ s rπ, 2 results in an element of C (q− ), that is a rational function in the variable q− , and therefore∧ has a meromorphic continuation to the entire plane, which we continue to denote as J (s,W,φ,µ,ψ). Similarly, we continue to denote the meromorphic continuation of J˜(s,W,φ,µ,ψ) by the same symbol. Furthermore, denote m I = spanC J (s,W,φ,µ,ψ) W (π, ψ) ,φ (F ) , π,ψ,µ { | ∈ W ∈ S } C 1 C s s then there exists a unique element p (Z) [Z], such that p (0) = 1 and Iπ,ψ,µ = p(q−s) [q− , q ]. ∈ 2 1 p (Z) does not depend on ψ, and we denote L (s, π, µ)= − . ∧ ⊗ p(q s) 6 RONGQING YE & ELAD ZELINGHER

Proposition 2.3 ([YZ18, Proposition 3.2]). 1) For n =2m, I X g J˜(s,W,φ,µ,ψ)= W σ m ψ ( trX) 2m I g Nm GLm(F ) m Mm(F ) m − Z \ ZB \ s 1 ι det g − µ (det g) φ (ε g ) dXd×g, ·| | Fψ 1 where ε1 = 1 0 ... 0 . 2) For n =2m +1, J˜(s,W,φ,µ,ψ)= Nm GLm(F ) M1×m(F ) m Mm(F ) Z \ Z ZB \ I X g I tZ 1 m m − W σ2m+1 Im g Im ·  I2m       1 1 1 s ψ (trX) det g µ (detg) φ (Z)dXdZd ×g,    · − | | Fψ Theorem 2.4 ([Mat14, Theorem 4.1] [CM15, Theorem 3.1]). There exists a non-zero 2 s m element γ (s, π, µ, ψ) C (q− ), such that for every W (π, ψ), φ (F ) we have ∧ ⊗ ∈ ∈ W ∈ S J˜(s,W,φ,µ,ψ)= γ s, π, 2 µ, ψ J (s,W,φ,µ,ψ) . ∧ ⊗ Furthermore, 2 1 L (1 s, π, µ− ) γ s, π, 2 µ, ψ = ǫ s, π, 2 µ, ψ − ∧ ⊗ , ∧ ⊗ ∧ ⊗ L (s, π, 2 µ) ∧ ⊗ 2 ks where ǫ (s, π, µ, ψ)= c q− , for k Z and c C∗. e ∧ ⊗ · ∈ ∈

The proof of the functional equation is very similar to the proofs of the referred theorems, and requires only slight modiﬁcations.

As before, we denote L (s, π, 2)= L (s, π, 2 1), γ (s, π, 2, ψ)= γ (s, π, 2 1, ψ), and ǫ (s, π, 2, ψ)= ǫ (s, π, 2 1, ψ∧). ∧ ⊗ ∧ ∧ ⊗ ∧ ∧ ⊗ Theorem 2.5. Suppose that (π,Vπ) is an irreducible supercuspidal representation of GLn (F )

1) If n =2m +1, then L (s, π, 2 µ)=1. 2 ∧ ⊗ 1 2) If n = 2m, then L (s, π, µ) = − , where p (Z) C [Z] is a polynomial, such ∧ ⊗ p(q s) ∈ that p (0) = 1 and p (Z) 1 ω (̟) µ (̟)m Zm. | − π

The proof of this statement is very similar to the proof of [Jo18, Theorem 6.2]. Its proof uses a slight modiﬁcation of [Jo18, Proposition 3.1] for the twisted Jacquet-Shalika integral. See also [JS90, Section 8, Theorem 1], [JS90, Section 9, Theorem 2] for the analogous global statements. −s 2 ks p1(q ) Lemma 2.6. Let n = 2m. Suppose that γ (s, π, µ, ψ) = c q− −(1−s) , where ∧ ⊗ · p2(q ) 1 1 c C∗, k Z, p1,p2 C [Z], such that p1 (0) = p2 (0) = 1 and p1 (Z) and p2 (q− Z− ) ∈ ∈ ∈ 2 1 2 1 1 don’t have any mutual roots. Then L (s, π, µ) = −s , L (s, π, µ− ) = −s , p1(q ) p2(q ) 2 ks ∧ ⊗ ∧ ⊗ ǫ (s, π, µ, ψ)= c q− . ∧ ⊗ · e EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 7

2 1 2 1 1 2 Proof. Write L (s, π, µ) = −s , L (s, π, µ− ) = −s , and ǫ (s, π, , ψ) = ∧ ⊗ pπ(q ) ∧ ⊗ pπe (q ) ∧ kπ s c q− · , where c C∗, k Z and p ,pe C [Z] satisfy p (0) = pe (0) = 1. π · π ∈ π ∈ π π ∈ π π e Then by Theorem 2.4 we have the equality s s 2 s k p1 (q− ) s kπ pπ (q− ) γ s, π, µ, ψ = c q− = cπ q− , 1 s 1 1 s 1 ∧ ⊗ · p2 q− (q− )− · pπe q− (q− )− which implies that k 1 1 kπ 1 1 cZ p1 (Z) pπe q− Z− = cπZ pπ (Z) p2 q− Z− , 1 as elements of the polynomial ring C [Z,Z− ].

m m 1 m m By Theorem 2.5, pπ (Z) 1 ωπ (̟) µ (̟) Z and pπe (Z) 1 ωπ− (̟) µ (̟)− Z . | − 1 1 | − Therefore, we get that pπ (Z) and pπe (q− Z− ) have no mutual roots. Note that they also don’t have zero or inﬁnity as a root. Therefore we conclude that every root of pπ (including multiplicity) is a root of p , which implies that p (Z) = h (Z) p (Z), where h C [Z], 1 1 1 π 1 ∈ with h1 (0) = 1. Similarly, we get that p2 (Z) = h2 (Z) pπe (Z), where h2 C [Z], with k kπ 1 1 ∈ 1 1 h2 (0) = 1. Hence we get that cZ h1 (Z)= cπZ h2 (q− Z− ). Since p1 (Z) and p2 (q− Z− ) don’t have any mutual roots, and since both don’t have zero or inﬁnity as a root, we get that h1 (Z) , h2 (Z) are constants. Therefore h1 = h2 = 1, and the result follows. q.e.d.

2.3. Simple supercuspidal representations. Let n be a positive integer.

Let ω : F ∗ C∗ be a multiplicative character such that ω ↾ = 1. → 1+p + 1 Let In = ν− (Nn (f)) be the pro-unipotent radical of GLn (F ), where Nn (f) is the upper + unipotent subgroup of GLn (f). Denote Hn = F ∗In .

Let t o∗/1+ p = f∗. Let t o∗ be a lift of t , i.e. ν (t)= t . We deﬁne a generic aﬃne 0 ∈ ∼ ∈ 0 0 character χ : H C∗ by n → n 1 − χ (zk)= ω (z) ψ ai + tan , i=1 ! X where z F ∗, and ∈ x1 a1 x a∗ ··· ∗  ∗ 2 2 ··· ∗  . . . . + k = ...... In .   ∈  xn 1 an 1  ∗ ∗ ··· − −  ̟an xn   ∗ ··· ∗  Note that χ does not depend on the choice t, because the conductor of ψ is p.

1 Let ζ C be an nth root of ω (t− ̟). ∈

In−1 j Denote g = − , H′ = g H . We deﬁne a character χ : H′ C∗ by χ (g h)= n t 1̟ n h ni n ζ n → ζ n ζjχ (h), for j Z and h H . ∈ ∈ n 8 RONGQING YE & ELAD ZELINGHER

ζ GLn(F ) Theorem 2.7 ([KL15, Section 4.3]). The representation σ = ind ′ (χζ ) is an irre- χ Hn ducible supercuspidal representation of GLn (F ).

ζ A representation σχ such as in Theorem 2.7 is called a simple supercuspidal represen- ζ tation. We say that π = σχ is a simple supercuspidal representation with central character ω, associated with the data (t0,ζ). By [KL15, Corollary 5.3], there exist exactly n (q 1) equivalence classes of simple supercuspidal representations with a given central character,− each of which corresponds to a pair (t0,ζ).

By [AL16, Section 3.3], we have that if W : GL (F ) C is the function supported on n → NnHn′ (where Nn is the upper triangular unipotent subgroup of GLn (F )), deﬁned by

W (uh′)= ψ (u) χ (h′) u N , h′ H′ , (2.1) ζ ∈ n ∈ n then W σζ , ψ is a Whittaker function. ∈ W χ 3. Computation of the twisted exterior square factors

In this section, we compute the twisted exterior square factors of a simple cuspidal representation. Throughout this section, let t0 o∗/1+ p = f∗, t o∗, ω : F ∗ C∗, ζ C∗ ζ ∈ ∼ ∈ → ∈ be as in Section 2.3. We denote π = σχ. Our goal is to compute the twisted exterior square factors of π.

3.1. Preliminary lemmas. In order to compute the twisted exterior factors of π, we will 1 use the function π σ2−m W , where W is the Whittaker function from Section 2.3. Before beginning our computation, we need some lemmas regarding the support of the integrand of 1 the twisted Jacquet-Shalika integral J s, π σ2−m W,φ,µ,ψ .

Im−l Im−l Denote for 1 l m, dl = −1 , wl = , and denote by τl the permuta- ≤ ≤ t ̟Il Il tion deﬁned by the columns ofwl, i.e.

wl = eτl(1) ... eτl(m) .

Lemma 3.1. Suppose that g GLm (f), X =(xij) m− (f) is a lower triangular nilpotent matrix, such that ∈ ∈N

Im X g 1 I2m 2l σ2m σ2−m N2m (f) − N2m (f) , Im g ∈ I2l for 1 l m. Then ≤ ≤

1) g Nm (f) wlNm (f). ∈ 1 1 2) If g wlNm (f), then xij =0 for every j < i such that τl− (j) < τl− (i), or equivalently ∈ + 1 + X m− (f) wl m (f) wl− , where m (f) is the subgroup of Mm (f) consisting of upper∈ N triangular∩ nilpotentN matrices. N EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 9

Furthermore, for g w N (f) and such X, ∈ l m Im X g 1 I2m 2l σ2m σ2−m = − v, Im g I l 2 where v N2m (f) is an upper triangular unipotent matrix, having zeros right above its diagonal.∈

Proof. The lemma is proved in [YZ18, Lemma 2.33] for the case that g = wdu, where w is a permutation matrix, d is a diagonal matrix and u N (f). Therefore we need only to show ∈ m the ﬁrst part for general g. By the Bruhat decomposition, we can write g = u1wdu2, where u1,u2 Nm (f), w is a permutation matrix, and d is a diagonal matrix. Denote g′ = wdu2. We have∈ 1 Im X g 1 u1 1 Im u1− Xu1 g′ 1 σ2m σ2−m = σ2m σ2−mσ2m σ2−m. Im g u Im g′ 1 u1 1 1 We have that σ ( u ) σ− N (f). Write u− Xu = L + U, where L − (f) is a 2m 1 2m ∈ 2m 1 1 ∈ Nm lower triangular nilpotent matrix and U m (f) is an upper triangular matrix. Then we have that ∈ B 1 Im u1− Xu1 Im U 1 Im L σ2m = σ2m σ2−mσ2m . Im Im Im Im U 1 u1 1 Since σ σ− N (f), and since σ ( u ) σ− N (f), we get that 2m Im 2m ∈ 2m 2m 1 2m ∈ 2m Im L g′ 1 I2m 2l σ2m σ2−m N2m (f) − N2m (f) . Im g′ ∈ I2l Since g′ = wdu2, we get from [YZ18, Lemma 2.33] that wd = wl, as required. q.e.d.

Lemma 3.2 ([YZ18, Lemma 2.34]). 1) Let d GLm (f) be a diagonal matrix. Then (m) (l) (l−m) ∈ Nm (f) wldNm (f) = q 2 − 2 − 2 Nm (f) . 2) |The set | ·| | + 1 1 1 − (f) w (f) w− = (x ) − (f) x =0, j < i s.t. w− (j)

Lemma 3.3. Suppose that a = diag (a1,...,am) is an invertible diagonal matrix, and X − (F ) is a lower nilpotent matrix, such that ∈Nm Im X a r σ2m = λ u g2m k, (3.1) Im a · · · where λ F ∗, u N , 1 r 2m, k K . Then ∈ ∈ 2m ≤ ≤ ∈ 2m 1) r =2l is even, for some 1 l m. ≤ ≤ 2) a1 = = am l = λ . | | ··· | − | | | 3) am l+1 = = am = λ ̟ . | 1− | ··· | | | |·| | 4) d− Xd − (o). l l ∈Nm 10 RONGQING YE & ELAD ZELINGHER

a 1 b = σ σ− = diag (a , a ,...,a , a ) . 2m a 2m 1 1 m m 2l Then buZσ2m = λug2mk. Let uZ = nZtZ kZ be an Iwasawa decomposition (nZ N2m, tZ A2m, kZ K2m). 1 1 1 2l 1 1 ∈ ∈ 1 1 ∈ Then we have λ− btZ = bn−Z b− u g2m kσ2−mkZ− . Denote u′ = bnZ− b− u N2m. Then we get ∈ I2m−2l − 1 1 −1 1 u′− λ− btZ K2m. (t ̟) I2l ∈ 1 1 Writing tZ = diag (t1,...,t2m), we get that λ − ai t2i = 1 and λ − ai t2i 1 = 1, 1 | | | || | 1 | | | || − | for every 1 i m l, and that λ − ai t2i = ̟ and λ − ai t2i 1 = ̟ , for every m l≤+1 ≤ i −m. By [JS90| ,| Section| || 5,| Proposition| | | 4],| t| || 1− for| odd| i| and − ≤ ≤ | i| ≥ ti 1 for even i. Thus we get that ti = 1 for every i. Hence, a1 = = am l = λ | | ≤ | | | | ··· | − | | | and am l+1 = = am = λ ̟ . | − | ··· | | | |·| | 4) By [JS90, Section 5, Proposition 5], there exists α > 0, such that if Z = (zij), then 1 α 1 max1 i,j m zij 1 i 2m ti . This implies that Z = a− Xa Mm (o). Since ≤ ≤ ≤i odd≤ | | ≤ | | m∈ ti = 1, we have a =Qλ dl k′, where k′ GLm (o) Am = (o∗) , and this implies | |1 · · ∈ ∩ d− Xd M (o), and therefore in − (F ) M (o)= − (o). l l ∈ m Nm ∩ m Nm q.e.d.

Lemma 3.4. Let g = ak, where a = diag (a1,...,am) is an invertible matrix, k GLm (o), 1 m ∈ d×g = δ− (a) d×k d×a , and X − (F ) be a lower triangular nilpotent matrix. If Bm i=1 i ∈Nm Q Im X g 1 σ2m σ2−m N2mH2′m, Im g ∈ Im X g 1 2l + then there exists 1 l m, such that σ2m Im ( g ) σ2−m F ∗N2mg2mI2m. Moreover, if Im X g ≤1 ≤ 2l + ∈ σ ( g ) σ− λN g I for 1 l m and λ F ∗, then 2m Im 2m ∈ 2m 2m 2m ≤ ≤ ∈

1 1 l(m l) 1) a = λd diag (u ,...,u ), where u ,...,u o∗, δ− (a)= δ− (d )= q− − . l 1 m 1 m ∈ Bm Bm l 2) Let k′′ = diag (u1,...,um) k. Then ν (k′′) Nm (f) wlNm (f), d×k = d×k′′. 1 ∈ 1 3) If ν (k′′) wlNm (f), then X = dlZdl− and dX = δB−m (dl) dZ, where Z m− (o) ∈ + 1 ∈ N satisﬁes ν (Z) − (f) w (f) w− . Moreover in this case, ∈Nm ∩ lNm l Im X g 1 2l σ2m σ2−m = λg2mv, Im g where v GL2m (o) satisﬁes ν (v) N2m (f), ν (v) has zeros right above its diagonal, and v has∈ zero at its bottom left corner.∈ EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 11

Proof. 1) Suppose that

Im X a k 1 r σ2m σ2−m = λug2mk′, (3.2) Im a k Z + r 1 where λ F ∗, u N2m (F ), r , k′ I2m. Since g2m = t− ̟I2m, we may assume ∈ ∈ ∈ ∈ 1 (by modifying λ) that 1 r 2m. By Lemma 3.3, we have that r =2l, X = d Zd− , ≤ ≤ l l where Z m− (o), and a = λdl diag (u1,...,um), for some u1,...,um o∗. ∈N · I2m−2l ∈ 2) Let k′′ = diag (u1,...,um) k and dl′ = −1 . Using these notations and part · t ̟I2l 1, we have that

Im X g 1 dl Im Z k′′ 1 σ2m σ2−m = λσ2m σ2−m. (3.3) Im g dl Im k′′ dl 1 Since d = σ σ− , we get from eq. (3.3) l′ 2m dl 2m

Im X g 1 Im Z k′′ 1 σ2m σ2−m = λdl′σ2m σ2−m. (3.4) Im g Im k′′ 2l I2m−2l Recall that r = 2l. Writing g = d′ , we get by combining eq. (3.2) and 2m l I2l eq. (3.4) that

Im Z k′′ 1 1 2l 1 I2m 2l σ2m σ2−m = dl′− ug2mk′ = dl′− udl′ − k′, (3.5) Im k′′ I l 2 1 I2m−2l which implies that d′− ud′ N2m (o), as ,k′ K2m, and the left hand side l l ∈ I2l ∈ 1 + + of eq. (3.5) is in K2m. Since dl′− udl′ N2m(o) I2mand k′ I2m, we get from eq. (3.5) that ∈ ⊆ ∈

Im Z k′′ 1 I2m 2l ν σ2m σ2−m N2m (f) − N2m (f) . Im k′′ ∈ I2l Since Z − (o), ν (Z) − (f), and by applying Lemma 3.1, we have that ν (k′′) ∈Nm ∈Nm ∈ Nm (f) wlNm (f). + 1 3) Assume that ν (k′′) wlNm (f), then by Lemma 3.1 we have ν (Z) m− (f) wl m (f) wl− , and ∈ ∈N ∩ N Im Z k′′ 1 I2m 2l ν σ2m σ2−m = − v′, Im k′′ I l 2 where v′ N2m (f) is an upper triangular matrix, having zeros right above its diagonal. Therefore∈ Im Z k′′ 1 I2m 2l σ2m σ2−m = − v, (3.6) Im k′′ I l 2 where v GL2m (o) satisﬁes ν (v)= v′. Combining eq. (3.6), eq. (3.4) and the fact that 2l ∈ I2m−2m g = d′ , we get 2m l I2l

Im X g 1 I2m 2l 2l σ2m σ2−m = λdl′ − v = λg2mv. Im g I l 2 Finally, suppose that l < m. Note that a non-zero scalar multiple of the last row of Im X g 1 v appears as the 2m 2l row of σ2m Im ( g ) σ2−m. The (2m 2l, 1) coordinate of Im X g 1− Im X g − σ m ( g ) σ− is the (2m l, 1) coordinate of ( g ), and this is zero, as 2 Im 2m − Im 12 RONGQING YE & ELAD ZELINGHER

2m 1 2m l > m. If l = m, then g2m = t− ̟I2m, and therefore the last row of v is a scalar − Im X g 1 multiple of the last row of σ2m Im ( g ) σ2−m, which has zero as its ﬁrst coordinate. q.e.d.

3.2. The even case. In this section, we compute the twisted exterior square factors for the even case n =2m.

Theorem 3.5. Let π be a simple supercuspidal representation of GL2m (F ) with central character ω, associated with the data (t0,ζ). Let µ : F ∗ C∗ be a unitary tamely ramiﬁed 2 m 1 1→ m C character, i.e. µ ↾1+p=1. Denote ξ = ζ µ ( 1) − t− ̟ . Let (ω µ )ξ : F ∗ ∗ be the m j j· − m · → character deﬁned by (ω µ ) (̟ u)= ξ ω (u) µ (u) , for j Z, u o∗. Then · ξ ∈ ∈ m 1 2 (s 1 ) − m γ s, π, µ, ψ = ξq− − 2 γ s, (ω µ ) , ψ . ∧ ⊗ · ξ Explicitly,

m 1) If (ω µ ) ↾ ∗ =1, then · o 6 m 1 2 (s 1 ) − 1 1 m γ s, π, µ, ψ = ξq− − 2 ψ (λ) ω λ− µ λ− . ∧ ⊗ q √ λ f∗ X∈ In this case L (s, π, 2 µ)=1, ǫ (s, π, 2 µ, ψ)= γ (s, π, 2 µ, ψ). m ∧ ⊗ ∧ ⊗ ∧ ⊗ 2) If (ω µ ) ↾ ∗ =1, then · o m 2 s 2 s 1 − 1 ξq− −( − 2 ) γ s, π, µ, ψ = ξq −1 (1 s) . ∧ ⊗ 1 ξ− q− − − 2 1 In this case L (s, π, µ)= 1 ξq−s , ∧ ⊗ − 2 m 2 (m 2)(s 1 ) ǫ s, π, µ, ψ = ξ − q− − − 2 . ∧ ⊗ Proof. We will compute the twisted exterior square gamma factor by computing the 1 ˜ 1 twisted Jacquet-Shalika integrals J s, π σ2−m W,φ,µ,ψ and J s, π σ2−m W,φ,µ,ψ , where W is the Whittaker function introduced in Section 2.3, and φ : F m C is the function de- ﬁned by → ψ ( ν (x )) x =(x ,...,x ) om, φ (x)= − 1 1 m ∈ . (3.7) (0 otherwise Then m 2 m q δε1 (ν (x)) x o , ψφ (x)= ∈ , F (0 otherwise where δ (x) is the indicator function of ε = (1, 0,..., 0) fm. ε1 1 ∈ 1 By the Iwasawa decomposition, we have that J s, π σ2−m W,φ,µ,ψ is given by EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 13

Im X ak 1 s J = W σ2m σ2−m det a µ (det (ak)) Im ak | | (3.8) Z 1 φ (εak) ψ ( trX) δ− (a) dXd×ad×k, · − Bm where W is the Whittaker function deﬁned in Section 2.3, given by eq. (2.1), X is integrated over M (F ), k is integrated over K = GL (o) and a is integrated over the diagonal Bm\ m m m subgroup Am of GLm (F ).

Denote by W the group of m m permutation matrices. By the Bruhat decomposition m × for GLm (f), we have the disjoint union

GLm (f)= Nm (f) wd0Nm (f).

w Wm ∈ d0 GAm(f) ∈ We decompose each of the double cosets of the disjoint union into a disjoint union of left cosets: given w W , d A (f), we can write ∈ m 0 ∈ m

Nm (f) wd0Nm (f)= u0wd0Nm (f),

u0 C ∈Gwd0 where C N (f) is a subset of N (f) such that the map wd0 ⊆ m m C u wd N (f) u N (f) , wd0 →{ 0 0 m | 0 ∈ m } u u wd N (f) 0 7→ 0 0 m is a bijection. We may assume without loss of generality that Im Cwd0 . We have that Nm(f)wd0Nm(f) ∈ C = | | . wd0 Nm(f) | | | | We obtain the following decomposition:

GLm (f)= u0wd0Nm (f).

w Wm u0 C ∈ ∈ wd0 d0 GAm(f) G ∈ 1 Since ν− (GLm (f)) = GLm (o), we can lift the above decomposition to

1 GLm (o)= uwdν− (Nm (f)),

w Wm u Cwd d∈GDm ∈G ∈ m where Dm Am GLm (o) = (o∗) is a set of representatives for the inverse image 1 ⊆ ∩ 1 ν− (A (f)) (i.e. D ν− (A (f)) and ν ↾ : D A (f) is a bijection), and for m m ⊆ m Dm m → m d Dm with ν (d) = d0, Cwd Nm (o) is a set of representatives for the inverse image ∈1 1 ⊆ ν− (C ) (i.e. C ν− (C ) and ν ↾ : C C is a bjiection). Without loss of wd0 wd ⊆ wd0 Cwd wd → wd0 generality, we may assume that the identity matrix belongs to Dm, and also belongs to Cwd, for every w W and d D . ∈ m ∈ m 14 RONGQING YE & ELAD ZELINGHER

Using this decomposition for Km in eq. (3.8), we decompose the integral J into a sum of integrals

J = Jwd,u,

w Wm u Cwd dX∈Dm X∈ ∈ where

Im X auwdk 1 s Jwd,u = W σ2m σ2−m det a µ (det (auwdk)) Im auwdk | | Z 1 φ (εauwdk) ψ ( trX) δ− (a) dXd×ad×k, · − Bm + 1 where X is integrated over m Mm (F ), k is integrated over Im = ν− (Nm (f)), and a is B \ 1 1 integrated over A . Writing au = aua− a, we have that aua− N , and since the m · ∈ m Jacquet-Shalika integrand is invariant under Nm GLm (F ), we have Jwd,Im = Jwd,u for any u C . Denote J = J , then we have \ ∈ wd wd wd,Im Nm (f) wν (d) Nm (f) J = Cwd Jwd = | |Jwd. | | Nm (f) w Wm w Wm | | dX∈Dm dX∈Dm ∈ ∈

Using the isomorphism M (F ) = −, we can write Bm\ m ∼ Nm Im X awdk 1 s Jwd = W σ2m σ2−m det a µ (det (awdk)) Im awdk | | (3.9) Z 1 φ (εawdk) δ− (a) dXd×ad×k, · Bm where the integration is the same as in J , except that this time X is integrated on −. wd,u Nm By Lemma 3.4, J = 0 unless w = w for some 1 l m. In this case by Lemma 3.4, wd l ≤ ≤ we have that the integrand of Jwld is supported on

a = λd diag (u ,...,u ) , where λ F ∗,u ,...,u o∗, l 1 m ∈ 1 m ∈ m d×a = d×λ d×ui, (3.10) i=1 1 1 Y l(m l) δB−m (a)= δB−m (dl)= q− − . Denote

k′′ = diag (u1,...,um) wldk. (3.11)

By Lemma 3.4, k′′ satisﬁes ν (k′′) N (f) w N (f). Since ν (k) N (f), then by the ∈ m l m ∈ m Bruhat decomposition of ν (k′′), we must have

ν (diag (u1,...,um)) wlν (d)= wl. Therefore, we have 1 1 diag (u1,...,um) = diag (u1′ ,...,um′ ) wld− wl− , (3.12) m m where u1′ ,...,um′ 1+ p and i=1 d×ui = i=1 d×ui′. Denote g = awldk. By eq. (3.10) and eq. (3.12), we have∈ Q Q g = awldk = λdldiag (u1′ ,...,um′ ) wlk. (3.13) EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 15

By eq. (3.11) and eq. (3.12), we have k′′ = diag (u′ ,...,u′ ) w k, and therefore ν (k′′) 1 m l ∈ wlNm (f). Hence, by part 3 of Lemma 3.4, 1 1 + 1 X = d Zd− , where Z ν− − (f) w (f) w− , l l ∈ Nm ∩ lNm l 1 l(m l) (3.14) dX = δB−m (dl) dZ = q− − dZ. Moreover, we have that

Im X g 1 2l σ2m σ2−m = λg2mv, (3.15) Im g where v GL2m (o) satisﬁes ν (v) N2m (f), ν (v) has zeros right above its diagonal, and v has zero∈ at its bottom left corner.∈ Therefore by eq. (2.1) and eq. (3.15), in this domain

Im X g 1 2l W σ2m σ2−m = ζ ω (λ) . (3.16) Im g

We have by eq. (3.13)

m m 1 l l(m l) det (aw dk)= λ t− ̟ u′ ( 1) − det k, (3.17) l i − i=1 ! Y and s ms ls det a = λ q− . (3.18) m | | | | Since det k 1+ p, i=1 ui′ 1+ p, and since µ is a tamely ramified character, and since 2 ∈ ∈ ( 1)l =( 1)l, we get by eq. (3.17) − − Q m m 1 1 l µ (det (aw dk)) = µ (λ) µ ( 1) − t− ̟ . (3.19) l − 1 m We have by eq. (3.13) that εawldk = um′ λt− ̟εlk. Since um′ 1+ p o∗, x F m m m ∈ ⊆ ∈ satisfies x o if and only if u′ x o . In the case x o , we have x u′ x (mod p), ∈ m ∈ ∈ ≡ m which implies that if x = (x ,...,x ) then x u′ x (mod p). Since ψ has conductor p, 1 m 1 ≡ m 1 ψ (um′ x1) = ψ (x1). Therefore, by the definition of φ in eq. (3.7), φ (um′ x) = φ (x) for every x F m. It follows that ∈ 1 1 φ (εawldk)= φ um′ λt− ̟εlk = φ λt− ̟εlk . (3.20) Substituting in eq. (3.9) the equalities eq. (3.10), eq. (3.12), eq. (3.14), eq. (3.16), eq. (3.18), eq. (3.19) and eq. (3.20), we get

2l ms ls m m 1 1 l J = ζ ω (λ) λ q− µ (λ) µ ( 1) − t− ̟ wld | | − Z m (3.21) 1 l(m l) l(m l) φ λt− ̟ε k q− − q− − dZ d×λ d×u′ d×k, · l i i=1 ! Y 1 + 1 where the integral is integrated over λ F ∗, u1′ ,...,um′ 1+p, Z ν− m− (f) wl m (f) wl− , k I+. ∈ ∈ ∈ N ∩ N ∈ m 16 RONGQING YE & ELAD ZELINGHER

2 m 1 1 Denote ξ = ζ µ ( 1) − t− ̟ . We can now evaluate the integration over Z,u1′ ,...,um′ in eq. (3.21) and get− + 1 2l(m l) ls m− (f) wl m (f) wl− 1 l Jwld =q− − q− N ∩ N m ξ · m− (f) f∗ |N | | | (3.22) m ms 1 ω (λ) µ (λ) λ φ λt− ̟εlk d×λd×k. · + ∗ | | ZIm ZF Notice that eq. (3.22) implies that Jwld does not depend on d Dm, and we have Jwl = Jwld for every d D . Denote ∈ ∈ m Nm (f) wlν (d) Nm (f) Jl = Cwld Jwld = | |Jwl . (3.23) | | Nm (f) d Dm d Dm X∈ X∈ | | By Lemma 3.2,

− Nm (f) wlν (d) Nm (f) + 1 (m) (l) (l m) | | = − (f) w (f) w− = q 2 − 2 − 2 . (3.24) N (f) Nm ∩ lNm l | m | m We also have Dm = Am (f) = f∗ . Therefore by substituting into eq. (3.23) the equalities eq. (3.22) and| eq.| (3.24),| we| have| |

(m) ls l m ms 1 Jl = q− 2 q− ξ ω (λ) µ (λ) λ φ λt− ̟εlk d×λd×k, (3.25) · + ∗ | | ZIm ZF (m) m l m l m where the expression q− 2 arises from the identity 2l (m l)+2 2 2 − = − − 2 − 2 − 2 − 2 m . − 2 + m Since k Im, we have that εlk o has 1 as its l coordinate modulo p. Therefore if 1 ∈ m ∈1 1 j λt− ̟εlk o , we must have λt− ̟ 1, i.e. λ = (t− ̟) u0, for some u0 o∗ and j 1. For∈ a ﬁxed k I+ we decompose| | ≤ · ∈ ≥ − ∈ m m ms 1 ω (λ) µ (λ) λ φ λt− ̟εlk d×λ F ∗ | | Z (3.26) ∞ 1 j 1 jm jms m 1 j+1 = ω t− ̟ µ t− ̟ q− ω (u0) µ (u0) φ u0 t− ̟ εlk d×u0 · ∗ j= 1 Zo X− m 2m m(m 1) 1 m 2m 1 Since ξ = ζ µ ( 1) − µ (t− ̟) , m (m 1) is even, and ζ = ω (t− ̟), we have − − m 1 1 m ξ = ω (t− ̟) µ (t− ̟) . Therefore, we get from eq. (3.26) that

m ms 1 ω (λ) µ (λ) λ φ λt− ̟εlk d×λ F ∗ | | Z (3.27) ∞ s jm m 1 j+1 = ξq− ω (u0) µ (u0) φ u0 t− ̟ εlk d×u0. · ∗ j= 1 Zo X− If l 2, then εlk has 0 as its ﬁrst coordinate modulo p, so for every j 1, the ﬁrst ≥ 1 j+1 1 j+1 ≥ − coordinate of u0 (t− ̟) εlk is 0 modulo p. Thus φ u0 (t− ̟) εlk = 1. We also have EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 17 that m m 1 (ω µ ) ↾o∗ =1 ω (u0) µ (u0) d×u0 = · . ∗ 0 otherwise Zo ( Therefore, from eq. (3.25) and eq. (3.27), we get for l 2 that, ≥ − m q ( 2 ) m l s s 1 m ∗ (ξq )− (ξq ) m −ms (ω µ ) ↾ =1 [GLm(f):Nm(f)] − − 1 ξ q o Jl = − · . 0 otherwise 

 1 j+1 If l = 1 and j 0, we have that (t− ̟) εlu0k 0 (mod p), and therefore we have again 1 j+1 ≥ ≡ φ u0 (t− ̟) εlk = 1 and

m m 1 (ω µ ) ↾o∗ =1 ω (u0) µ (u0) d×u0 = · . ∗ 0 otherwise Zo ( If l = 1 and j = 1, we have that ε1k has 1 as its ﬁrst coordinate modulo p, and therefore 1 j+1 − φ u (t− ̟) ε k = ψ ( ν (u )), and we have 0 l − 0 m 1 m ω (u0) µ (u0) ψ ( u0) d×u0 = ω (λ) µ (λ) ψ ( λ). ∗ − f − o ∗ λ f∗ Z | | X∈ To summarize, we get:

m − −s m+1 q ( 2 ) m (ξq ) ξq−s s m ∗ (ξq− )− m −ms (ω µ ) ↾o =1, [GLm(f):Nm(f)] 1 ξ q − q 1 · J1 = − − .  − m q ( 2 ) s m s 1 m  (ξq− )− (ξq− ) ∗ ∗ ω (λ) µ (λ) ψ ( λ) otherwise  [GLm(f):Nm(f)] f λ f | | ∈ − Summing all the Jl up, we get P − m m q ( 2 ) (m 2) 1 ξ−1q−(1−s) s m ∗ (ξq− )− − q − −s (ω µ ) ↾o =1, [GLm(f):Nm(f)] · · (1 ξq )(q 1) · J = Jl =  − m − − . q ( 2 ) s (m 1) 1 m l=1  (ξq− )− − ∗ ∗ ω (λ) µ (λ) ψ ( λ) otherwise  [GLm(f):Nm(f)] f λ f X | | ∈ − (3.28)  P 1 We now move to compute J˜ = J˜ s, π (σ2m)− W,φ,µ,ψ . Following the same steps as before for the expression in Proposition 2.3, we have m

J˜ = J˜l, l X=1 where

m m m(s 1) ˜ ( 2 ) l(s 1) l 1 1 ι Jl = q− q− − ξ ω (λ) µ (λ) λ − ψφ λ− ε1dl− wlk d×λd×k. · + ∗ | | F ZIm ZF Recall that m 2 m q δε1 (ν (x)) x o , ψφ (x)= ∈ . F (0 otherwise 18 RONGQING YE & ELAD ZELINGHER

We have that 1 ι 1 1 ι λ− εl+1k 1 l m 1 λ− ε1dl− wlk = 1 1 1 ι ≤ ≤ − . (λ− (t− ̟)− ε1k l = m 1 1 ι ι m If 1 l m 1, we have that λ− ε1dl− wlk is a scalar multiple of εl+1k o and the l +1 ≤ ≤ − ι 1 1 ι ∈ coordinate of εl+1k is 1 modulo p, and therefore λ− ε1dl− wlk can not be in the support of 1 1 ι ι m ψφ for any scalar λ. If l = m, λ− ε1dm− wlk is a scalar multiple of ε1k o , which satisﬁes F ι 1 1 ι m ∈ ν (ε1k ) ε1 (mod p). Therefore, in order for λ− ε1dl− wlk to be in o and to be ε1 modulo ≡ 1 1 1 p, we must have l = m, and λ− (t− ̟)− 1+ p. Hence we have J˜l =0 for 1 l m 1, ∈ 1 1 ≤ ≤ − and for l = m, we have that λ is integrated on (t− ̟)− (1 + p) and that

(m) m(s 1) m 1 1 1 m 1 m(s 1) m J˜m = q− 2 q− − ξ ω t− ̟ − µ t− ̟ − t− ̟ − − q 2 d×λd×k · + ZIm Z1+p (m) 1 q− 2 m = q 2 . f [GL (f): N (f)] · | ∗| m m Therefore (m) 1 q− 2 m J˜ = J˜ = q 2 . (3.29) m f [GL (f): N (f)] · | ∗| m m m Recalling the fact that when (ω µ ) ↾ ∗ = 1, the Gauss sum · o 6 G (ω µm, ψ)= ω (λ) µ (λ)m ψ ( λ) · − λ f∗ X∈ has absolute value √q, we have that

m m 1 G (ω µ , ψ) 1 1 m G (ω µ , ψ)− = · = ω λ− µ λ− ψ (λ). (3.30) · q q λ f∗ X∈ By eq. (3.28), eq. (3.29), and eq. (3.30) we get

m 2 s 1 1 ξq−s ( 2 ) − m ∗ ˜ ξq− − −− − −s (ω µ ) ↾o =1, 2 J 1 ξ 1q (1 ) γ s, π, µ, ψ = = m 1 · − · . ∧ ⊗ J  (s 1 ) − 1 1 m − − 2 ∗ ξq √q λ f ω (λ− ) µ (λ− ) ψ (λ) otherwise  ∈ m P1 2  (s 1 ) − m The formula γ (s, π, µ, ψ)= ξq− − 2 γ s, (ω µ ) , ψ now follows from a stan- ∧ ⊗ · ξ dard computation of the local factors of Tate’s functional equation, see for instance [RV99, Section 7.1] or [Kud03, Proposition 3.8].

The claim about the other twisted exterior square factors now follows from Lemma 2.6 1 1 1 and the fact that 1 ξZ and 1 ξ− q− Z− don’t have mutual roots. q.e.d. − −

3.3. The odd case. In this section, we compute the twisted exterior square factors for the odd case n =2m + 1. EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 19

Theorem 3.6. Let π be a simple supercuspidal representation of GL2m+1 (F ) with central character ω, associated with the data (t ,ζ). Let µ : F ∗ C∗ be a unitary tamely ramiﬁed 0 → character, i.e. µ ↾1+p=1. Then m 2 1 2 (s 1 ) γ s, π, µ, ψ = µ t− ̟ ζ q− − 2 . ∧ ⊗ Furthermore, in this caseL (s, π, 2 µ)=1, ǫ(s, π, 2 µ, ψ)= γ (s, π, 2 µ, ψ). ∧ ⊗ ∧ ⊗ ∧ ⊗

1 ˜ 1 Proof. We compute J s, π σ2−m+1 W,φ,µ,ψ and J s, π σ2−m+1 W,φ,µ,ψ , where again W is the Whittaker function introduced in Section 2.3, but this time m m m δ0 (ν (x)) x o , q− 2 x o , φ (x)= ∈ , ψφ (x)= ∈ , (0 otherwise F (0 otherwise where δ is the indicator function of 0 fm. 0 ∈ 1 By the Iwasawa decomposition, J s, π σ2−m+1 W,φ,µ,ψ is given by

Im X ak Im 1 J = W σ2m+1 Im ak Im σ2−m+1 φ (Z)         (3.31) Z 1 1 Z 1 s 1 1 det a − µ (det (ak)) δ− (a) dXd×ad×kdZ,   ·| | Bm where X is integrated on −, a is integrated on the diagonal matrix subgroup A , k is Nm m integrated on Km = GLm (o), and Z is integrated on M1 m (F ). In order for Z to be in the × support of φ, we must have Z M1 m (o), such that ν (Z) = 0, i.e. Z M1 m (p). For such ﬁxed Z, we have that ∈ × ∈ ×

Im 1 ν σ2m+1 Im σ2−m+1 = I2m+1,   Z 1  and therefore     Im 1 + σ2m+1 Im σ2−m+1 I2m+1.  Z 1 ∈ Hence, in order for X, a, k to contribute to the integral, we must have

Im X ak 1 l σ2m+1 Im ak σ2−m+1 = λu′g2m+1k′,  1  1  Z  + 2m+1 1 where λ F ∗, u′ N2m+1, l , k′ I2m+1. Since g2m+1 = t− ̟I2m+1, we may assume (by modifying∈ λ), that∈ 1 l 2m∈ +1. Notice∈ that ≤ ≤ Im X ak l 1 1 λg2m+1k′ = u′− σ2m+1 Im ak σ2−m+1, (3.32)  1  1     and the right hand side of eq. (3.32) has ε2m+1 = (0,..., 0, 1) as its last row. On the l 1 other hand, the last row of λg2m+1k′ is λt− ̟εlk′, where εl is the l-th standard row vector. 20 RONGQING YE & ELAD ZELINGHER

Since εlk′ is the l-th row of k′, we have that ν (εlk′) is the l-th row of an upper triangular 1 unipotent matrix, and therefore the equality λt− ̟εlk′ = ε2m+1 can’t hold unless l =2m+1. Thus we have l = 2m +1, and that the last row of k′ is a scalar multiple of ε2m+1. Since k′ GL2m+1 (o), we may assume (by modifying λ by a unit) that the last row of k′ is ε2m+1. ∈ k′′ v + We write k′ = ( 1 ), where k′′ I2m and v is a column vector in M2m 1 (o). Writing ′′ ′′ × I2m v k ∈ k k′ = 1 ( 1 ), we may assume (by modifying u′) that k′ = ( 1 ), which implies that u′′ 1 u′ =( ) for u′′ N . Thus we get that λt− ̟ = 1, and that 1 ∈ 2m Im X ak 1 2m 2m + σ2m σ2−m = λu′′g2mk′′ λN2mg2m I2m. (3.33) Im ak ∈ 1 Since eq. (3.33) holds, we can apply Lemma 3.4 and use λt− ̟ = 1 to get that

a = λ d diag (u ,...,u ) = diag (u ,...,u ) , where u ,...,u o∗, · m · 1 m 1 m 1 m ∈ m

d×a = d×ui, (3.34) i=1 1 Y δB−m (a)=1. Denote

k0 = diag (u1,...,um) k, (3.35) d×k = d×k0. 1 Then k = diag (u1,...,um)− k0, and by Lemma 3.4 1 1 + k ν− (N (f) w N (f)) = ν− (N (f)) = I (3.36) 0 ∈ m m m m m Furthermore, since ν (k0) Nm (f) = wmNm (f), by Lemma 3.4 we have that X m− (o) ∈ + 1 ∈ N and that X satisﬁes ν (X) − (f) (w (f) w− )= 0 , i.e. ∈Nm ∩ mNm m { m} X − (p) . (3.37) ∈Nm Also, since ν (k ) w N (f), by Lemma 3.4 we have for such Z, X, a, k that 0 ∈ m m Im X ak 1 σ2m σ2−m = v0, Im ak where ν (v0) = v0′ is an upper triangular unipotent matrix having zeros right above its diagonal, and therefore

Im X ak Im 1 Y = σ2m+1 Im ak Im σ2−m+1  1  1  Z 1 (3.38)       Im v0 1 = σ Im σ− 1 2m+1   2m+1 Z 1 ′   v0 satisﬁes ν (Y ) = 1 , which implies that ν (Y ) is an upper unipotent matrix with zeros right above its diagonal. Since Y also has zero at its left bottom corner, we have from eq. (2.1) W (Y )=1. (3.39) EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 21

+ From eq. (3.34) and eq. (3.35), we have that ak = k0 Im, so det (ak) = det (k0) 1+ p, which implies ∈ ∈

s det a =1, (3.40) | | µ (det (ak))=1, as µ is tamely ramiﬁed.

Therefore, we have by substituting in eq. (3.31) the equations (3.34)-(3.40) that m 1 J s, π σ2−m+1 W,φ,µ,ψ = dX d×ui d×k0dZ + ∗ m − M1×m(p) Im (o ) m (p) i=1 ! Z Z Z ZN Y (3.41) 1 1 1 = . M1 m (f) [GLm (f): Nm (f)] m− (f) | × | |N | ˜ 1 We now move to compute J s, π σ2−m+1 W,φ,µ,ψ . By Proposition 2.3, we need to evaluate the integral t Im X ak Im Z ˜ 1 − 1 J = W σ2m+1 Im ak Im σ2−m+1  I2m        Z 1 1 1 s 1 φ (Z) det a µ (det (ak)) δ− (a)dXd×ad×kdZ,    ·Fψ | | Bm (3.42) where again X m−, a Am, k Km = GLm (o), Z M1 m (F ). We notice that for every ∈N ∈ ∈ ∈ × Z M1 m (o), ∈ × t Im Z − 1 + σ2m+1 Im σ2−m+1 I2m+1.  1  ∈ Therefore, in order for X, a, k to support the integrand, we need

Im X ak 1 1 l σ2m+1 Im ak σ2−m+1 = λu′g2m+1k′, (3.43) I2m     1 1  Z +  where λ F ∗, u′ N2m+1, l , k′ I2m+1. Note that the left hand side of eq. (3.43) ∈t ∈ ∈ ∈ l has e1 = (1, 0,..., 0) as its last column. Therefore, λu′g2m+1k′ needs to have e1 as its l l 1 last column, i.e. λu′g2m+1k′e2m+1 = e1, which implies λg2m+1k′e2m+1 = u′− e1 = e1. Since l the 2m +1 l coordinate of ν g2m+1k′e2m+1 is 1, we must have l = 2m. Therefore, from 2m − λg2m+1k′e2m+1 = e1 we get that the last column of k′ is a scalar multiple of e2m+1. Modifying k′′ λ by a unit, we may assume that k′ has e2m+1 as its last column. Write k′ = ( v 1 ), where + 2m 1 k′′ I2m, v M1 2m (p). Writing g2m+1 = t−1̟I , we have ∈ ∈ × 2m 2m 1 k′′ λu′g2m+1k′ = λu′ 1 t− ̟I m v 1 2 1 1 1 v (t− ̟k′′)− 1 k′′ = λu′ 1 . I t− ̟I2m 1 2m 22 RONGQING YE & ELAD ZELINGHER

− ′′ −1 1 v(t 1̟k ) k′′ Therefore, by replacing u′ by u′ , we may assume k′ = ( 1 ), which implies I2m 1 by eq. (3.43) that u′ =( u′′ ) foru′′ N2m. Substituting the expressions for u′, k′, and the 2m 1 ∈ −1 expression g2m+1 = t ̟I2m in eq. (3.43), we get that

Im X ak 1 1 λt− ̟u′′k′′ σ Im ak σ− = , 2m+1     2m+1 λ 1 1 and therefore λ = 1 and   

Im X ak 1 1 σ2m σ2−m = t− ̟u′′k′′. Im ak By Lemma 3.4, we have that 1 a = d diag (u ,...,u )= t− ̟diag (u ,...,u ) , where u ,...,u o∗, m · 1 m 1 m 1 m ∈ m

d×a = d×ui, (3.44) i=1 1 Y δB−m (a)=1.

Denote

k0 = diag (u1,...,um) k, (3.45) d×k0 = d×k. Then by Lemma 3.4, 1 1 + k ν− (N (f) w N (f)) = ν− (N (f)) = I . (3.46) 0 ∈ m m m m m Since ν (k0) wmNm (f) = Nm (f), by Lemma 3.4 we have X Mm (o) satisﬁes ν (X) ∈ + 1 ∈ ∈ − (f) (w (f) w− )= 0 , i.e. Nm ∩ mNm m { m} X − (p) . (3.47) ∈Nm Also, since ν (k ) w N (f), by Lemma 3.4 we have for such elements that 0 ∈ m m Im X ak 1 2m 1 σ2m σ2−m = g2mv0 = t− ̟v0, Im ak where v0 GL2m (o) satisﬁes that ν (v0)= v0′ is an upper triangular unipotent matrix with zeros right∈ above its diagonal. Hence, we have that t Im X ak Im Z 1 − 1 Y = σ2m+1 Im ak Im σ2−m+1 I2m       1 1 1 (3.48)    t    Im Z 2m v0 − 1 = g σ Im σ− . 2m+1 1 2m+1   2m+1 1 Denote   t Im Z v0 − 1 Y ′ = σ Im σ− , 1 2m+1   2m+1 1   EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 23

2m then Y = g2m+1Y ′, and ν (Y ′) is an upper triangular unipotent matrix with zeros right above its diagonal, as it is a product of such. Y ′ also has zero at its left bottom corner. Therefore by eq. (2.1) W (Y )= ζ2m. (3.49) 1 1 m By eq. (3.44) and eq. (3.45), we have that ak = t− ̟k0, and therefore det (ak)=(t− ̟) det k0. Since k I+, then det k 1+ p, which implies 0 ∈ m 0 ∈ s ms det a = q− , | | 1 m (3.50) µ (det (ak)) = µ t− ̟ , as µ is tamely ramiﬁed. By substituting in eq. (3.42) the equations (3.44)-(3.50), we have m 2m m ms 1 m J˜ = ζ q− 2 q− µ t− ̟ dX d×ui d×k0dZ + ∗ m − M1×m(o) Im (o ) m (p) Z Z Z ZN i=1 ! (3.51) Y 1 1 m 2m ms 1 m = q− 2 ζ q− µ t− ̟ . [GL (f): N (f)] (f) m m |Nm− | We get from eq. (3.41) and eq. (3.51), ˜ 2 J 1 m 2m ms m γ s, π, µ, ψ = = µ t− ̟ ζ q− q 2 . ∧ ⊗ J The result regarding the other local factors now follows from Theorem 2.4 and Theo- rem 2.5. q.e.d.

4. Exterior square gamma factors local converse theorem

In this section we present and prove a local converse theorem for simple supercuspidal representations. Unlike previous local converse theorems, which are usually based on Rankin- Selberg gamma factors, our theorem is based on twisted exterior square gamma factors.

Theorem 4.1. Let n =2m or n =2m +1. Let (π,Vπ), (π′,Vπ′ ) be simple supercuspidal representations of GLn (F ), with the same central character ω = ωπ = ωπ′ , such that π, n 1 π′ are associated with the data (t0,ζ) and (t0′ ,ζ′) correspondingly, where ζ = ω (t− ̟), n 1 ζ′ = ω t′− ̟ , and t, t′ o∗ are lifts of t0, t0′ respectively, i.e. ν (t) = t0, ν (t′) = t0′ . Assume that ∈ 1) If n =2m, then gcd(m 1, q 1)=1. 2) If n =2m +1, then gcd(−m, q − 1)=1. −

Suppose that for every unitary tamely ramiﬁed character µ : F ∗ C∗, we have → 2 2 γ s, π, µ, ψ = γ s, π′, µ, ψ . (4.1) ∧ ⊗ ∧ ⊗ Then ζ = ζ′ and t = t′ . ± 0 0 24 RONGQING YE & ELAD ZELINGHER

Proof. Suppose n = 2m. Let µ be a unitary tamely ramiﬁed character. Denote ξ = 2 m 1 1 2 m 1 1 ζ µ ( 1) − t− ̟ , ξ′ = ζ′ µ ( 1) − t′− ̟ . We claim that ξ = ξ′. · − · − m If (ω µ ) ↾ ∗ = 1, we have by Theorem 3.5 and by eq. (4.1) that · o m 2 s m 2 s (s 1 ) − 1 ξq− (s 1 ) − 1 ξ′q− ξq− − 2 − = ξ′q− − 2 − , (4.2) 1 (1 s) 1 (1 s) 1 ξ− q− − 1 ξ′− q− − − − and we get ξ = ξ′ by comparing the poles of both sides of eq. (4.2).

m If (ω µ ) ↾ ∗ = 1, we have by Theorem 3.5 and by eq. (4.1) that · o 6 m 1 m 1 (s 1 ) − 1 1 m (s 1 ) − 1 1 m ξq− − 2 ψ (λ) ω λ− µ λ− = ξ′q− − 2 ψ (λ) ω λ− µ λ− . q q √ λ f∗ √ λ f∗ X∈ X∈ Therefore m 1 ξ − =1. (4.3) ξ ′ On the other hand, 2 m 1 1 − 2 ξ ζ µ ( 1) t− ̟ ζ 1 = − = µ t′t− . (4.4) 2 m 1 1 2 ξ′ ζ µ ( 1) − t − ̟ ζ′ ′ − ′ 2m 1 2m 1 Since ζ = ω (t− ̟) and ζ′ = ω(t′− ̟), we get from eq. (4.4) that m 1 ξ ω (t− ̟) 1 m 1 1 m = µ t′t− = ω t′t− µ t′t− , ξ ω (t 1̟) ′ ′− which implies that m(q 1) ξ − =1, (4.5) ξ ′ as ω,µ are tamely ramiﬁed. Since m 1 is coprime to m and to q 1, we have that gcd(m 1, m (q 1)) = 1, and therefore−m 1 is invertible modulo m (q − 1), which implies − − ξ − − from eq. (4.3) and eq. (4.5) that ξ′ = 1, i.e. ξ = ξ′.

1 1 2 We proved ξ = ξ′. Therefore by eq. (4.4), we have µ (t′t− )=(ζ′ζ− ) for every unitary 2 2 tamely ramified character µ. Choosing the trivial character, this implies that ζ = ζ′ . 1 Suppose t = t′ , then there exists a unitary character µ : f∗ C∗, such that µ t′ t− = 1, 0 6 0 0 → 0 0 0 6 and we can lift this to a unitary tamely ramified character µ : F ∗ C∗ that satisfies µ ↾o∗ = 1 1 → 1 1 2 µ ν and then µ (t′t− )= µ t′ t− = 1, which is a contradiction to µ (t′t− )=(ζ′ζ− ) = 1. 0 ◦ 0 0 0 6 Therefore, we must have t = t′ . 0 0 For n =2m + 1, the proof is similar. We have that 2m+1 1 ζ ω (t− ̟) 1 = = ω t′t− , ζ ω (t 1̟) ′ ′− and therefore (2m+1)(q 1) ζ − =1, (4.6) ζ ′ EXTERIOR SQUARE γ-FACTORS FOR SIMPLE SUPERCUSPIDAL REPRESENTATIONS 25 as ω is tamely ramified. By eq. (4.1) and Theorem 3.6, we have for any unitary tamely ramified character µ m m 1 2 (s 1 ) 1 2 (s 1 ) µ t− ̟ ζ q− − 2 = µ t′− ̟ ζ′ q− − 2 , which implies that 2m ζ 1 m = µ t′− t . (4.7) ζ ′ 2m 2m Substituting the trivial character in eq. (4.7), one gets ζ = ζ′ , which implies that ζ2 m =1. (4.8) ζ 2 ′ Since gcd (m, q 1) = gcd(m, 2m +1) = 1, we get that gcd(m, (2m +1)(q 1)) = 1, and − − ζ2 therefore m is invertible modulo (2m +1)(q 1). By eq. (4.6), eq. (4.8), this implies ′ = 1. − ζ 2 By eq. (4.7) and eq. (4.8), we have for every unitary tamely ramified character µ : F ∗ C∗, 1 m 1 q 1 → µ t′− t = 1. Since µ (t′− t) − = 1, and since m is coprime to q 1, m is invertible modulo − 1 q 1, and therefore we have that for every unitary tamely ramified character µ, µ t′− t = 1. − As in the even case, this implies t = t′ . q.e.d. 0 0 2 2 2m+1 Remark 4.2. In the odd case, we actually get that ζ = ζ′, since ζ = ζ′ and ζ = 2m+1 1 ζ′ = ω (t− ̟), and then 2m+1 2m+1 ζ ζ′ ′ ζ = 2 m = 2 m = ζ . (ζ ) (ζ′ )

As a consequence, when the hypotheses in Theorem 4.1 are met, we have t0 = t0′ and ζ = ζ′, so π ∼= π′.

Acknowledgments. The authors thank Zahi Hazan for introducing them to simple supercuspidal representations. The authors thank James Cogdell for his comments on an earlier version of this paper.

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Department of Mathematics, Purdue University, West Lafayette, IN 47907, USA

E-mail address: [email protected]

Department of Mathematics, Yale University, New Haven, CT 06510, USA

E-mail address: [email protected]