A I Badulescu

A La

Virgen Del Rocío

On p-adic Speh representations

This note contains simple proofs of some known results (unitarity, character formula) on Speh representations of a group GL n (D) where D is a local non Archimedean division algebra of any characteristic.

Introduction. In this note I give simple proofs of some known results on Speh representations of groups GL n (D) where D is a central division algebra of finite dimension over a local non archimedean field F of any characteristic. The new idea is to use the Moeglin-Waldspurger algorithm (MWA) for computing the dual of an irreducible representation. For the unitarizability of Speh representations, previous proofs were based either on the trace formula and the close fields theory, or on deep results of type theory. The proof here is "combinatoric", independent of D, the characteristic, and Bernstein's (also called U0) theorem. Short general proofs using MWA are also given for other known facts. The proof is always the same: one wants to prove a relation (R) involving some representations (for example an induced representation π is irreducible). One starts by writing the "naive" relation (R') between these representations, known from standard theory, but not as strong as (R) (for example the semi simplification of π is a sum with non negative coefficients of some irreducible representations). Usually (R') has more terms that (R), because it is weaker, and one wants to prove that some terms which are supposed to appear in (R') are actually not there (for example all the subquotients of π except of the expected one have coefficient zero). The method then is to consider also the dual relation (R") to (R'), and to play with the MWA, in order to show that the mild constraints one has on (R') and (R") are enough to show that the extra terms are not there and (R') reduces actually to (R).

All the results in this paper are already known, and here I only give new short proofs. So I kindly ask the reader, when using one of these facts, to quote, at least in first place, the original reference (see the historical notice at the end of the paper). Beside the Zelevinsky and Tadić classification of the admissible dual ( [Ze], [START_REF] Tadić | Induced representations of GL(n; A) for a p-adic division algebra A[END_REF]), the proofs here rely on [Au] (dual of an irreducible representation is irreducible), [MW] and [START_REF] Badulescu | Zelevinsky involution and Moeglin-Waldspurger algorithm for GL(n, D)[END_REF] (algorithm for computing the dual) and some easy tricks from [START_REF] Tadić | Classification of unitary representations in irreducible representations of a general linear group (non-Archimedean case)[END_REF], [START_REF] Badulescu | Global Jacquet-Langlands correspondence, multiplicity one and classification of automorphic representations Invent[END_REF] and [CR], and do not involve any complicated technique.

The idea of searching for "simple proof" for classification of unitary representations, together with a list of basic tricks to use, are due to Marko Tadić ([Ta3] for example). He also was the first to formulate some properties of Speh representations (formula for ends of complimentary series, character formula, Speh representations are prime elements in the ring of representations, dual of a Speh representation is Speh) and to prove them when D = F . The starting point of my proof here of the assertion Speh representations are unitary is also due to Tadić who found the simple but brilliant trick reducing the problem of unitarity to a problem of irreducibility.

I would like to thank Guy Henniart who read the paper and made useful observations. π ⊗ π . The results in this section for which no other reference is given may be found in [Ze] (for D = F ) and [START_REF] Tadić | Induced representations of GL(n; A) for a p-adic division algebra A[END_REF] (for general D). In [START_REF] Tadić | Induced representations of GL(n; A) for a p-adic division algebra A[END_REF] the characteristic zero is assumed, but this restriction may be removed ([BHLS] 2.2). Let Gr n denote the Grothedieck group of admissible representations of finite length of GL n (D), and, by convention, Gr 0 = Z. Set R := ⊕ n∈N Gr n . The composition law × induces a composition law * in R which makes R(+, * ) into a commutative Z-algebra (× is not commutative, but * is). We write representation for isomorphy class of representations where no confusion may occur. The set Irr of irreducible representations of GL n (D), n ∈ N × , is a natural linear basis of R.

Let δ be an essentially square integrable (i.e. twist of a square integrable by a character) representation of GL n (D). Let s(δ) be the smallest positive real number x such that δ×ν x δ is reducible. Then s(δ) ∈ N * and s(δ)|d (so s(δ) = 1 if D = F ). Set ν δ := ν s(δ) . Then δ may be obtained as the unique irreducible subrepresentation of a representation of type

ν a δ ρ × ν a-1 δ ρ × ν a-2 δ ρ × ... × ν a-m+1
δ ρ, with a ∈ R, m a divisor of n, and ρ a cuspidal unitary representation of GL n m (D). Also, a, m and ρ are determined by δ. We then write δ = Z(ρ, a, m). We have s(ρ) = s(δ). We set e(δ) := a -m-1 2 . Then δ is unitary (i.e. is square integrable) if and only if e(δ) = 0. In this case we simply write δ = Z(ρ, m),

understood as: δ is the unique subrepresentation of ν m-1 2 δ ρ × ν m-1 2 -1 δ ρ × ν m-1 2 -2 δ ρ × ... × ν -m-1 2 δ ρ.
We make the convention that Z(ρ, a, 0) is the unit element of R.

To any irreducible representation π of GL n (D) we may associate a unique multiset [δ 1 , δ 2 , ..., δ k ] such that δ i is an essentially square integrable representations of some GL n i (D), n i ≥ 1, i n i = n and -if we reorder the d i such that e(δ i ) is decreasing with i, then π is the unique irreducible quotient of δ 1 × δ 2 × ... × δ k .

We write π = L([δ 1 , δ 2 , ..., δ k ]). We call standard representation an element of R which is a product δ 1 * δ 2 * ... * δ k with d i essentially square integrable representations. It is known that δ 1 * δ 2 * ... * δ k determines the multiset [δ 1 , δ 2 , ..., δ k ], so we may write π = L(δ 1 * δ 2 * ... * δ k ) instead of π = L([δ 1 , δ 2 , ..., δ k ]). Let D be the set of essentially square integrable representations in R and Std be the set of standard representations. Then L realizes a bijection from Std to Irr (and its restriction to D is the identity). If Θ is a standard representation and π = L(Θ), we then say that π is the Langlands quotient of Θ.

If δ is square integrable and k ∈ N * , then u(δ, k) (resp. ū(δ, k)) will denote the Langlands quotient of ν k-1 2 δ δ × ν k-3 2 δ δ × ν k-5 2 δ δ × ... × ν -k-1 2 δ δ (resp. of ν k-1 2 δ × ν k-3 2 δ × ν k-5 2 δ × ... × ν -k-1 2 δ). The representations u(δ, k) are called Speh representations.
If δ = Z(ρ, l), then δ + (and δ -) will stand for Z(ρ, l + 1) (and Z(ρ, l -1)).

Let ι be the Aubert involution on R from [Au] (see also [ScS]) which we will call here duality. It takes an irreducible representation of GL n (D) to plus or minus an irreducible representation of GL n (D). We will denote i the involution of the set of irreducible representations induced by ι (i.e. forgetting the sign; for D = F it is the Zelevinsky involution from [Ze]). We recall that ι(π) * ι(π ) = ι(π * π ).

We call segment a subset of R of the form S = {a, a + 1, a + 2, ..., a + m -1}, m ∈ N, with the convention S = ∅ if m = 0. Set l(S) = m and call it the length of S. We say a is the beginning of S and a + m -1 is the ending of S.

We call multisegment a non empty multiset M = [S 1 , S 2 , ..., S k ] of non empty segments. Set m(M ) := max S∈M l(S) (the maxlength of M ) and t(M ) := k (the thickness of M ). Let B(M ) (resp. E(M )) be the multiset of the beginnings (resp. endings) of segments in M . We call support of M the multiset reunion of all the segments in M . Two segments S 1 and S 2 are said to be linked if S 1 ⊆ S 2 , S 2 ⊆ S 1 and S 1 ∪ S 2 is a segment. If M and M are multisegments we say M is obtained from M by an elementary operation if there exists a pair of linked segments

S 1 , S 2 in M such that M is obtained from M by replacing [S 1 , S 2 ] with [S 1 ∪ S 2 , S 1 ∩ S 2 ]
where it is understood we forget S 1 ∩ S 2 in case it is empty. One introduces then a partial order on multisegments by putting M ≤ M if M is obtained from M through a finite number of elementary operations. It is easy to see that on the partially ordered set of multisegments m is decreasing, while t, B and E are increasing (for the relation on sets

X ≤ Y if X ⊂ Y ). Let R be the set of unitary cuspidal representations in R. If ρ ∈ R, let K ρ be the set {ν α ρ ρ, α ∈ R}. Let Irr ρ (Std ρ , D ρ ) be the subset of Irr (Std, D) made of representations with cuspidal support included in K ρ . If ρ, ρ ∈ R and ρ = ρ , if π ∈ Irr ρ and π ∈ Irr ρ , then π * π ∈ Irr ([Ze], [Ta2]). Let R ρ be the linear span of Irr ρ in R. Then R ρ is a subalgebra of R and one has R = × ρ∈R R ρ . Each R ρ is stable by the involution ι.
Let us fix ρ ∈ R and for the rest of this paragraph consider only representations in R ρ . Then essentially square integrable representations are parametrized by segments and so standard representations are parametrized by multisegments. The (partial) order relation defined on the set of multisegments induces an order relation on Std ρ . The bijection L : Std → Irr induces a bijection from Std ρ to Irr ρ and we get by transfer a partial order on Irr ρ . We will use this bijection L to compare for example an irreducible representation to a standard representation. Also we will use E(π), t(π) etc. for an irreducible representation π, being understood that this is E(M ), t(M ) etc., where M is the multisegment associated to the standard representation Θ such that π = L(Θ). For any standard representation Θ, one has in R (actually in R ρ )

Θ = L(Θ) + π j <L(Θ) a i π j , a i ∈ N * , π j ∈ Irr (
this is an avatar of the Langlands quotient theory explicitly stated in this stronger form in [Ze] and [START_REF] Tadić | Induced representations of GL(n; A) for a p-adic division algebra A[END_REF] Theorem 5.3).

As a consequence, Std ρ is a linear basis of the sub-space R ρ . So (R ρ , +, * ) is isomorphic with the ring of polynomials over Z with (commutative) variables in the set D ρ . A standard representation Θ ∈ R ρ is then a monomial. Notice that its degree equals t(L(Θ)) and is also ≥ t(π j ) for all the π j < L(Θ) (because t is decreasing). Decomposing an irreducible representation on the standard basis comes to write it as a polynomial.

Important fact. If π ∈ R ρ has a decomposition π = i m i π i with m i = 0 and π i distinct irreducible representations, then, as polynomials, we have deg(π) = max i deg(π i ). To see this one writes π i = L(Θ i ) and notice that the degree of π i is also the degree of the monomial Θ i . If, among the π i of maximal degree as polynomials, π i 0 is maximal for the order relation on representations, then the monomial Θ i 0 appears with non zero coefficient in π, which shows that deg(π) ≥ max i deg(π i ). The inequality ≤ is obvious. Let M be a multisegment. We associate to M the multisegment M # in the following way : let e be the greatest number in E(M ). Chose:

-a shortest segment δ 0 ∈ M with ending e.

-a shortest segment δ 1 ∈ M with ending e -1 and not included in δ 0 -a shortest segment δ 2 ∈ M with ending e -2 and not included in δ 1 etc.. Stop when it is not possible to find the next segment. Say δ r is the last one. Then S 0 = {er, er + 1, ..., e} is the first segment of M # . Let now M -be the multisegment obtained from M after replacing δ i with δ - i for all i ∈ {0, 1, ..., r} (if one of these is empty, just erase it). Starting with M - what we have done with M , we find the second segment of M # , and so on (so in the end we have that M # is the multiset union of [S 0 ] and (M -) # ). As observed in [MW], if the greatest number in E(M ) is e, then all the segments of M # ending with e will be constructed exclusively with points from E(M ). As an easy consequence, we have the following property:

(P) If E(M ) is included in some segment S of length k (i.e. ∀x ∈ E(M ), x ∈ S), then l(M # ) ≤ k. (Indeed, all the segments of M # ending with the greatest point e of S will be included in S, so will have length ≤ k. And once we are done with e, the set of endings will be (as a consequence) included in the shift of S with -1.) Also, every segment of M # being constructed taking elements on distinct segments of M , we obviously have (P') : l(M # )) ≤ t(M ), i.e. the maxlength of M # is less than or equal to the thickness of M .

The results. In the sequel we will consider elements π of Irr which will actually be, most of the time, obviously in some Irr ρ . For these, we will loosely say things like "let M be the multisegment of π", or apply MWA, being understood we parametrized by segments. The assertions (b), (c), (d) in the following theorem are the (U1), (U2), (U3) of [START_REF] Tadić | Induced representations of GL(n; A) for a p-adic division algebra A[END_REF].

Theorem 1. Let δ be a square integrable representation of some GL n (D).

α ∈] -1 2 , 1 2 [. (d) u(δ, k) is a prime element of the ring R.
Proof of (a): is obvious by MWA.

Proof of (b): Tadić ([Ta1]) reduced it by induction to the proof :

(i) for all k ≥ 1, u(δ, k) × u(δ, k) is irreducible and (ii) for all k ≥ 2, u(δ, k -1) × u(δ, k + 1) is irreducible.
Here I prove (i) and (ii) (this trick of Tadić was also used in [BHLS] 4.1):

Proof of (i): let δ = Z(ρ, l) and set τ := Z(ρ, k). Write

u(δ, k) * u(δ, k) = π + π j <π a j π j
where π is the Langlands quotient of the product and a j ∈ N. The same way:

u(τ, l) * u(τ, l) = π + π j <π b j π j .
By (a), u(τ, l) is the dual of u(δ, k). The second relation is then dual to the first. The MWA implies that the Langlands quotient π is the dual of π. If some a j = 0, then π j appears effectively in the first sum and its dual, which we call π j , appears in the second sum by linear independence of the characters. As l(τ ) = k and π j < π , we have m(π j ) > k (at the first elementary operation the longest segment will already be longer than k; then m is decreasing). This is impossible by

(P), since E(π j ) ⊆ E(u(δ, k) * u(δ, k)) (E is increasing) which is included in a segment of length k. So a j = 0 for all j.
Proof of (ii): write

u(δ, k -1) * u(δ, k + 1) = π + π j <π a j π j
where π is the Langlands quotient of the product and a j ∈ N. Then write the dual relation (using (a) for the left side) :

u(τ -, l) * u(τ + , l) = π + π j <π b j π j
where MWA implies that the Langlands quotient π is the dual of π. Let π j , π j be two dual representations appearing effectively in the sums. As l(δ) = l we know m(π j ) ≥ l + 1. Now

E(π j ) ⊂ E(u(τ + , l) * u(τ -, l)) = x + [0, 1, 1, 2, 2, 3, 3, ..., l -1, l -1, l] and B(π j ) ⊂ B(u(τ + , l) * u(τ -, l)) = y + [0, 1, 1, 2, 2, 3, 3, ..., l -1, l -1, l]
where for the sake of the reader I shifted with x = k-l-1 2 and y = -k-l+1

2

. Notice that x + l always belongs to E(π j ), so if M is the multisegment of π j , then the construction of (the first segment of) M # starts with x + l. Now I will show that the first segment S 0 of M # (i.e. the one ending in x + l) does not contain x. Indeed, the segment δ 0 of π j containing x + l cannot be shorter than the one of u(τ + , l) * u(τ -, l) containing x + l (elementary operations cannot shorten this one). So the beginning of δ 0 is ≤ y + l -1. But the sequence of beginnings of segments δ 0 , δ 1 , δ 2 , ... used in the construction of S 0 has to be strictly decreasing, and there is no place in B(π j )\{y + l} for such a sequence with l + 1 terms. Hence the first segment of M # is constructed only with points from x + {1, 2, ..., l} so its length is ≤ l. But, as this one does not contain x, we have that E(M -) is included (in the sense of (P)) in the segment x + {0, 1, 2, ..., l -1}, which implies by (P) that the other segments of M # are also of length ≤ l. This is in contradiction with l(π j ) ≥ l + 1.

Proof of (c): Tadić showed in [START_REF] Tadić | Induced representations of GL(n; A) for a p-adic division algebra A[END_REF] irreducibility for α ∈] -1 2 , 1 2 [\{0} and also that unitarity would follow from irreducibility in α = 0, which has been proved in (i) (see the proof of (b)).

Proof of (d): Assume u(δ, k) is not a prime element of R. Then it is not a prime element of R ρ (easy) and we will work in this ring. As R ρ is a polynomial ring over Z, it is a UFD, hence

u(δ, k) is reducible. Write u(δ, k) = S * T
where S and T are non scalar elements of R ρ . Then

u(τ, l) = ι(S) * ι(T ).
As a polynomial, u(δ, k) has degree k and so deg(S), deg(T ) < k. Then, if we write S, T as linear combination of irreducible representations, every irreducible representation π in S or T has t(π) < k. So every irreducible representation i(π) in ι(S) or ι(T ) has m(i(π)) < k (by

(P')). But, if α is an irre- ducible representation in ι(S), then t(α) ≤ deg(ι(S)) and if β is an irreducible representation in ι(T ) then t(β) ≤ deg(ι(T )) so finally t(α) + t(β) ≤ l (because deg(ι(S)) + deg(ι(T )) = deg(u(τ, l)) = l).
But, as all the segments of α and β are strictly shorter than k, α * β lives on a group of smaller size than u(τ, l). It is impossible for u(τ, l) to be a sum of such representations.

Corollary 1. Let T be the set of all finite products of (irreducible unitary) representations of type u(δ, k) and π(u(δ, k), α) for k ∈ N * , δ square integrable and α ∈]0, 1 2 [. Let U be the set of all irreducible unitary representations of all GL m (D), m ∈ N * . Then T ⊂ U .

Proof. By prop. 2.13 in [START_REF] Badulescu | Global Jacquet-Langlands correspondence, multiplicity one and classification of automorphic representations Invent[END_REF], a product u 1 × u 2 × ... × u k of irreducible unitary representations is irreducible unitary if, for all 1 ≤ i ≤ k, u i × u i is irreducible. By (i) (see the proof of (b)), U = u(δ, k) × u(δ, k) is indeed irreducible. To show that π(u(δ, k), α) × π(u(δ, k), α) is also irreducible we write it as ν α δ U × ν -α δ U . Now ν α δ U and ν -α δ U are irreducible representations, and for 0 < α < 1 2 no segment of the first one is linked to no segment of the second one. So ν α δ U × ν -α δ U is irreducible by Proposition 2.2 and Lemma 2.5 of [START_REF] Tadić | Induced representations of GL(n; A) for a p-adic division algebra A[END_REF].

If δ = Z(ρ, l) is a square integrable representation and k ∈ N * , set

F (δ, k) := ν -k+l 2 ( w∈W l k (-1) sgn(w) k i=1 Z(ρ, i, w(i) + l -i)) ∈ R,
where W l k is the set of permutation w of {1, 2, ..., k} such that w(i) + l ≥ i for all i ∈ {1, 2, ..., k}.

Theorem 2.

If δ = Z(ρ, l) then one has, in R, u(δ, k) = F (δ, k).

(f ) One has ν

-1 2 δ u(δ, k) * ν 1 2 δ u(δ, k) = u(δ, k -1) * u(δ, k + 1) + u(δ -, k) * u(δ + , k).
Proof of (e): the case k = 1 is trivial, k = 2 is known by [Ze] and [START_REF] Tadić | Induced representations of GL(n; A) for a p-adic division algebra A[END_REF]. Assume by induction that u(δ, k) = F (δ, k) and u(δ, k -1) = F (δ, k -1) for any square integrable representation δ.

Let us write b j π j .

ν -1 2 δ u(δ, k) * ν 1 2 δ u(δ, k) = u(δ, k-1) * u(δ, k+1)+u(δ -, k) * u(δ + , k)+
These formulas need some explanation: as u(δ, k-1) * u(δ, k+1) is irreducible by (ii) (see the proof of

(b)), it is the Langlands quotient of ν -1 2 δ u(δ, k) * ν 1 2
δ u(δ, k). For the same reason, u(τ, l-1) * u(τ, l+1) is the Langlands quotient of ν -1 2 δ u(τ, l) * ν 1 2 δ u(τ, l). Now, as i(u(τ, l -1) * u(τ, l + 1)) = u(δ -, k) * u(δ + , k), this representation appears with multiplicity one in ν

-1 2 δ u(δ, k) * ν 1 2 δ u(δ, k) by duality. As i(u(δ, k -1) * u(δ, k + 1)) = u(τ -, l) * u(τ + , l), this representation appears with multiplicity one in ν -1 2 δ u(τ, l) * ν 1 2
δ u(τ, l) by duality. So the two Σ are dual to one another and we also have the property

[A]: the coefficient b j of π j = u(τ -, l) * u(τ + , l) is zero.
Chenevier and Renard ( [CR]) recognized in F (δ, k) a determinant with entries in R and proved using Lewis Carroll's identity that

[F] ν -1 2 δ F (δ, k) * ν 1 2 δ F (δ, k) = F (δ, k-1) * F (δ, k+1)+F (δ -, k) * F (δ + , k).
Using the induction assumption we replace to obtain:

ν -1 2 δ u(δ, k) * ν 1 2 δ u(δ, k) = u(δ, k -1) * F (δ, k + 1) + u(δ -, k) * u(δ + , k).
Subtracting from our first formula we get:

π j <u(δ,k-1) * u(δ,k+1) a j π j = u(δ, k -1) F (δ, k + 1) -u(δ, k + 1) .
So taking the dual, u(τ -, l) divides π j <u(τ,l-1) * u(τ,l+1) b j π j . Write π j <u(τ,l-1) * u(τ,l+1) b j π j = u(τ -, l) * c p θ p with c p ∈ Z * and θ p distinct irreducible representations. Chose a maximal θ p . Then L(u(τ -, l) * θ p ) is one of the π j such that b j = 0. But t(π j ) ≤ 2l so t(θ p ) ≤ l. As E(π j ) is included in a segment of length k + 1, m(π j ) ≤ k + 1 (by (P)) and so m(θ p ) ≤ k + 1. As the support of θ p has l(k + 1) elements and we proved it contains less then l segments, all of length less than k + 1, the only possibility is θ p = u(τ + , l). So u(τ -, l) * θ p = u(τ -, l) * u(τ + , l). As this representation is irreducible, we have L(u(τ -, l) * θ p ) = u(τ -, l) * u(τ + , l).This is in contradiction with [A]. So what we called c p θ p is void, and this implies (e) at level k + 1.

Proof of (f ): The formula [F] implies that (e) for all k implies (f ) ( [CR]).

Corollary 2. Let C be the Jacquet-Langlands transfer for square integrable representations from [DKV] between GL nd (F ) to GL n (D) and LJ the transfer of irreducible representations from [Ba2] between GL ndk (F ) to GL nk (D). If δ is a square integrable representation of GL nd (F ), then LJ(u(δ, k)) = ū(C(δ), k).

Proof. It is proved in [START_REF] Tadić | Representation theory of GL(n) over a p-adic division algebra and unitarity in the Jacquet-Langlands correspondence[END_REF] to be an easy consequence of the character formula (e). Notice the [DKV] transfer is known also when ch(F ) = 0 ([Ba1]).

Remark. Bernstein's theorem asserts that induced representations from unitary irreducible representations remain irreducible. Bernstein proved this theorem for D = F in [Be]. Scherre proved it for general D in [Se]. Tadić showed ([Ta1]) using Bernstein's theorem that actually T = U (the Corollary 2 here is just one inclusion) and this is the only proof of the equality we have up to now. The proof of the Bernstein's theorem is difficult, especially for general D, and a new short one would be most welcome. This theorem has not been used in this paper.

Historical notice. -Tadić gave then a proof of all the other results in this paper in [START_REF] Tadić | Representation theory of GL(n) over a p-adic division algebra and unitarity in the Jacquet-Langlands correspondence[END_REF] (2006) assuming only Bernstein's theorem were true for D = F (but not involving anymore Zelevinsky's derivatives which are specific to D = F ). I gave myself a proof of these facts in [START_REF] Badulescu | Global Jacquet-Langlands correspondence, multiplicity one and classification of automorphic representations Invent[END_REF] (2008) without using the Bernstein's theorem which was not known for D = F at that time. Eventually Scherre proved Bernstein's

Notation.

  Let (F, | | F ) be a local non archimedean field of any characteristic ch(F ) and D a central division algebra of finite dimension d 2 over F . Let ν n : GL n (D) → C × be the character g → | det(g)| F , denoted simply ν when no confusion may occur (det is the reduced norm). All representations here will be admissible of finite length. If π, π are representations of GL a (D) and GL b (D) such that a + b = n, let P be the parabolic subgroup of GL n (D) containing the upper triangular matrices and of Levi factor GL a (D) × GL b (D) and set π × π := ind GLn(D) P

  Using the relation R = × ρ∈R R ρ one may see that Std is a linear basis of R and R is a polynomial algebra with variables in D. Every element of Std may be written in a unique way as a product of elements of Std ρ , ρ ∈ R, which are almost all trivial. So we get a product order relation on Std. This order relation passes then to Irr.If Θ ∈ Std, one has then in R Θ = L(Θ) + π j <L(Θ) a i π j with a i ∈ N * and π j ∈ Irr. And if Θ = ρ∈A Θ ρ where A is a finite subset of R and Θ ρ ∈ Std ρ ,then L(Θ) = ρ∈A L(Θ ρ ). Let π, π ∈ Irr and set π = L(Θ), π = L(Θ ) with Θ, Θ ∈ Std. Then π * π = x + π j <x a i π j ,where x = L(Θ * Θ ), a i ∈ N, π j ∈ Irr. We will call x the Langlands quotient of π * π .The algorithm. For this section we fix ρ ∈ R and we parametrize Irr ρ by multisegments. Let π ∈ Irr ρ . If M and M # are the multisegments of π and i(π), Moeglin and Waldspurger gave an algorithm for computing M # from M ([MW] for D = F , [BR2] for D = F ). I will call here MWA this algorithm. Here is a description: If δ = {b, b + 1, ..., e} is a non empty segment, we set δ -= {b, b + 1, ..., e -1} (which may be empty).

  (a) If δ = Z(ρ, l) then i(u(δ, k)) = u(τ, l) where τ := Z(ρ, k). (b) The representation u(δ, k) is unitary. (c) The representations π(u(δ, k), α) := ν α δ u(δ, k) × ν -α δ u(δ, k) are irreducible and unitary for

π

  j <u(δ,k-1) * u(δ,k+1) a j π j and taking the dual: τ, l) = u(τ, l-1) * u(τ, l+1)+u(τ -, l) * u(τ + , l)+ π j <u(τ,l-1) * u(τ,l+1)

For

  D = F , Theorem 1 and Corollary 1 have been proved by Tadić in [Ta1] (1986), (f ) in [Ta7] (1987), and (e) in [Ta4] (1995). Tadić used Bernstein's theorem ([Be]) and Zelevinsky derivatives. He conjectured this results should hold also for D = F in [Ta2] (1990), where he proved (d) in this case. For D = F , ch(F ) = 0: -(b) has been proved in [BR1] (2004) for D of characteristic zero (as it requires a result from [Ba2] based on the Arthur trace formula).