Conjectures and results about parabolic induction of representations of GLn(F)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\text {GL}}_n(F)$$\end{document}

In 1980 Zelevinsky introduced certain commuting varieties whose irreducible components classify complex, irreducible representations of the general linear group over a non-archimedean local field with a given supercuspidal support. We formulate geometric conditions for certain triples of such components and conjecture that these conditions are related to irreducibility of parabolic induction. The conditions are in the spirit of the Geiss–Leclerc–Schröer condition that occurs in the conjectural characterization of □\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\square $$\end{document}-irreducible representations. We verify some special cases of the new conjecture and check that the geometric and representation-theoretic conditions are compatible in various ways.


Introduction
Let F be a local non-archimedean field. The smooth, complex representations of GL n (F), n ≥ 0 were studied in depth in the seminal work of Bernstein and Zelevinsky [4,5,31]. In particular, Zelevinsky obtained a classification of the set Irr of irreducible representations of GL n (F), n ≥ 0 in terms of multisegments-an essentially combinatorial object. Denote by Z (m) the irreducible representation corresponding to a multisegment m. A basic property is that for any π i = Z (m i ), i = 1, . . . , k, the representation Z (m 1 + · · · + m k ) occurs with multiplicity one in the Jordan-Hölder sequence of the representation π 1 × · · · × π k parabolically induced from π 1 ⊗ · · · ⊗ π k . In particular, if π 1 × · · · × π k happens to be irreducible, then it is equivalent to Z (m 1 + · · · + m k ). However, the problem of characterizing the irreducibility of Z (m 1 ) × · · · × Z (m k ) was left open. In this paper we study the case k = 2.
An important special case is when π 2 = π 1 . We say that a representation π is -irreducible if π × π is irreducible. It is a well-known consequence of a theorem of Bernstein [7] that every unitarizable representation π ∈ Irr is -irreducible. However, our emphasis in this paper is on general irreducible representations. The first example of a non--irreducible π ∈ Irr was given by Leclerc [22].
The analogues of -irreducible representations in different (but related) contexts (where a different terminology is used) play an important role in the monoidal categorification of cluster algebras, a problem considered by Hernandez-Leclerc [11,12] and more recently by Kang-Kashiwara-Kim-Oh [15]. We will not say anything about this problem here except to mention that the argument of [14], adapted to the case at hand, shows that if π isirreducible, then for any π ∈ Irr, the socles soc(π × π ) and soc(π × π) are irreducible and each one occurs with multiplicity one in the Jordan-Hölder sequence of π × π [20]. It follows that if at least one of π = Z (m) and π = Z (m ) is -irreducible, then the irreducibility of Z (m) × Z (m ) is equivalent to the conjunction of the condition and its symmetric counterpart (interchanging m and m ).
On the other hand, in [29] Zelevinsky proposed a geometric framework for his classification. The purpose was twofold: to interpret multiplicities of irreducible representations in a standard module using Kazhdan-Lusztig theory, and to give a geometric interpretation of the Zelevinsky involution. The first goal was expounded in [30] and is best explained by the Arakawa-Suzuki functors [1]-see [20, §10] and the references therein for more details. The second goal was accomplished by Moeglin-Waldspurger who also give a combinatorial description of the resulting involution on multisegments [24]. (The fact that the Zelevinsky involution preserves irreducible representations is a special case of a more general result proved independently by Aubert [2,3] and Schneider-Stuhler [26]-see also [6] for a more recent treatment).
In this geometric framework, the work of Geiss-Leclerc-Schröer [9] suggests a simple, conjectural geometric criterion for the -irreducibility of Z (m). More precisely, fix a finite-dimensional graded vector space V over C and let G(V ) be the group of grading preserving automorphisms of V . Then, the multisegments with supercuspidal support determined by the graded dimension of V , parameterize the finitely many G(V )-orbits on the space of linear transformations A : V → V that are homogeneous of degree 1 (or alternatively, −1). They also parameterize the irreducible components of the variety X(V ) of commuting pairs of linear transformation A, B : V → V of degree 1 and −1 respectively. Denote by C m irreducible component corresponding to m. The conjecture of [20], inspired by [10,Conjecture 18.1], is that Z (m) is -irreducible if and only if C m admits an open G(V )-orbit. (See Sect. 4 for more details.) This condition can be explicated fairly concretely and can be efficiently checked, at least by a randomized algorithm-see [20] where also a special case of this conjecture was proved, and was related to singularities of Schubert varieties of type A.
The main goal of this paper is to propose and substantiate a conjectural geometric criterion for the irreducibility of Z (m) × Z (m ), provided that at least one of Z (m) and Z (m ) is -irreducible. More precisely, suppose that C m ⊂ X(V ) and C m ⊂ X(V ). Then, C m+m ⊂ X(V ⊕ V ) and we can embed C m × C m diagonally in C m+m .

Conjecture 1 Suppose that at least one of Z (m) and Z (m ) is -irreducible.
Then, More generally, under the above assumption we give a conjectural geometric criterion for (1). (See Sect. 5 for more details.) Recall that C m comes with a distinguished G(V )-invariant Zariski open subset C • m (containing the interior of C m in X(V ), but not equal to it in general). Let Y be the subvariety of C m+m consisting of the pairs (A, B) satisfying the following two conditions.

A(V ), B(V ) ⊂ V , and (A
The induced pair on the quotient V belongs to C • m .

Conjecture 2 Suppose that at least one of Z (m) and Z (m ) is -irreducible.
Then, This conjecture implies Conjecture 1. The geometric conditions in the two conjectures are in the spirit of the Geiss-Leclerc-Schröer condition. They are equally concrete and can be checked (by a randomized algorithm) very efficiently on a computer.
We will give supportive evidence for Conjecture 2, ergo, Conjecture 1. (See Sect. 6 for more details.) For instance, we show that they hold (without restriction on m ) in any one of the following cases.
1. Z (m) is unitarizable. 2. m is a ladder multisegment in the sense of [18]. 3. Z (m) is generic, or more generally every segment in m has the same length. 4. Z (m) is unramified, or more generally, the Zelevinsky dual of Z (m) is generic.
A key ingredient for the proof is the results of [19]. In fact, the conjecture was forged as an attempt to explain the results of [19] geometrically. We also show that Conjecture 2 satisfies a number of non-trivial consistency checks. While in all likelihood new ideas will be needed to establish the conjecture in general, we believe that the attestation that we already have so far cannot be coincidental.
In the best-case scenario, Conjecture 2 may in fact hold without restriction on m and m . If true, this would imply that However, we feel that at this stage it would be too parlous to postulate such a strong form of Conjecture 2 (even in one direction) since our evidence for this generality is indirect and nowhere near conclusive. The main reason is that we do not have a practical way to check the condition (1) independently in general. At any rate, the condition that Z (m + m ) is a direct summand of Z (m) × Z (m ) is strictly weaker than the irreducibility of Z (m) × Z (m ). In fact, there are plenty of examples of irreducible representations π of GL n (F) such that π × π is semisimple of length two. Consequently, Conjecture 1 would have been false had we lifted the assumptions on m and m . In order to obtain a precise irreducibility criterion without restriction on m and m , we would need to characterize the condition (which in general is stronger than (1)) and at present we do not have a conjectural geometric criterion for this.
In a different direction, a natural follow-up question, which we hope to address in the future, is to obtain a geometric insight on soc(Z (m) × Z (m )), assuming as before that at least one of Z (m) and Z (m ) is -irreducible.
Finally, we point out that our conjectures do not seem to lie in the scope of the Langlands program.
The contents of the paper are as follows. After introducing the relevant notation and the Zelevinsky classification (Sect. 2) we recall the notion of -irreducible representations and basic facts about irreducibility of parabolic induction (Sect. 3). In Sect. 4 we recall the geometric condition of Geiss-Leclerc-Schröer and the conjecture relating it to -irreducibility. The heart of the paper is Sect. 5 where we state Conjectures 1 and 2 and analyze the pertinent geometric conditions. In Sect. 6 we state results confirming these conjectures in special cases and provide several consistency checks for them. These results are proved in Sect. 8 using Jacquet module techniques and other combinatorial tools which are recalled in Sect. 7.

General notation
Throughout the paper we fix a non-archimedean local field F with normalized absolute value |·|. For a non-negative integer n let M n denote the C-linear, locally finite, abelian category of complex, smooth, finite length (hence admissible) representations of the group GL n (F). Set In particular, write Irr M 0 = {1}. We denote by Irr c M n the subset of irreducible supercuspidal representations. By abuse of notation we often write π ∈ M to mean that π is an object of M.
As customary, normalized parabolic induction with respect to standard (block upper triangular) parabolic subgroup will be denoted by ×. This is a bilinear biexact bifunctor with associativity constraints given by induction in stages. In other words, × endows M with the structure of a ring category with unit element 1. The Grothendieck group R = ⊕ n≥0 R n of M inherits a structure of a graded commutative ring.
For any π ∈ M n and a character χ of F * , we denote by π · χ the representation obtained from π by twisting by the character χ • det. In particular, we write We denote by π ∨ the contragredient of π and by soc(π ) the socle of π , i.e., the largest semisimple subobject of π . (If π = 0, then soc(π ) = 0.)
Note that if π ∈ M is SI and 0 = τ → π then τ is SI and soc(τ ) = soc(π ). By the argument of [24, p. 173], for any π 1 , π 2 , τ ∈ Irr we have For any set A we denote by N(A) the free commutative monoid generated by the elements of A. It consists of finite formal sums of elements of A. The standard order on N(A) will be denoted by ≤.
We may view JH as a map from the objects of M to N(Irr). Thus, for τ ∈ Irr and π ∈ M we have τ ≤ π if and only if τ ≤ JH(π ) in N(Irr).

Zelevinsky classification and involution [31]
A segment (of length k) is a nonempty finite set of the form For compatibility, we also set We say that Δ, ) By definition, a multisegment is an element of N(Σ). We extend the maps Δ → supp(Δ) and Δ → Δ ∨ to an additive map supp : N(Σ) → N(Irr c ) and an additive involution m → m ∨ on N(Σ), respectively. Given m = Δ 1 + · · · + Δ k ∈ N(Σ) we may enumerate the Δ i 's (in possibly more than one way) such that Δ i ⊀ Δ j whenever i < j. Then, the representation z(m) = Z (Δ 1 ) × · · · × Z (Δ k ) is SI and up to equivalence, depends only on m. The main result of Zelevinsky is that the map is a bijection between N(Σ) and Irr. There is also a dual bijection given by m → L(m) := soc(λ(m)) where λ(m) = L(Δ k ) × · · · × L(Δ 1 ) (under the same assumption on the order of the Δ i 's). The latter is essentially the Langlands classification in this context. We refer the reader to [19, §3] for a summary of the basic properties of these bijections. In particular, Z (0) = L(0) = 1, s(Z (m)) = s(L(m)) = supp m and The two bijections are related by Deligne-Lusztig type duality. Let t denote the duality functor on M defined by Schneider-Stuhler in [26], composed with the contragredient (in order to make it covariant). (See also [6] for a more recent approach.) Then for all m ∈ N(Σ) and (π × π ) t = π t × π t for any π, π ∈ M.
Finally, let be the bijection (in fact, involution) such that L(m) = Z (m # ). Note that # is not additive. We recall the recursive algorithm, due to Moeglin-Waldspurger [24], for the computation of the involution m → m # . For convenience (although it is not absolutely necessary) we fix a total order ≤ on Irr c subject only to the condition that ρ < → ρ for all ρ ∈ Irr c . Set 0 # = 0. For 0 = m = i∈I Δ i ∈ N(Σ) define max m = max i∈I e(Δ i ).
Let m ≥ 1 and i 1 , . . . , i m ∈ I be indices such that max m = e(Δ i 1 ) and b(Δ i 1 ) is maximal with respect to this property (8a) There is the segment of m # such that e(Δ(m)) = max m = max m # and b(Δ(m)) is minimal with respect to this property.
and similarly for SSA(m, m ). A simple sufficient condition for SSA(m, m ) is the following (see e.g., [19,Proposition 3.5]).
SSA(m, m ) ⇐ ∀ segment Δ in m and Δ in m we have Δ ⊀ Δ . (10) 1 The notation m − should not be confused with the same notation in [31].

(11b)
As we will recall below, it is possible that Z (m) × Z (m ) is semisimple but not irreducible. In particular, SSA(m, m ) is strictly stronger than SA(m, m ).
The following key definition is inspired by the work of Hernandez-Leclerc and Kang-Kashiwara-Kim-Oh. [14,15,20]) An object π = 0 of M (necessarily in Irr) is called -irreducible if the following equivalent conditions are satisfied.
We denote by Irr ⊂ Irr the class of -irreducible representations. Clearly, this class is invariant under contragredient and the duality t . The first example of an irreducible representation π (of GL 8 (F)) which is not -irreducible was given by Leclerc [22]. Explicitly, let and for simplicity we write [a, b] for the segment {|·| a , . . . , |·| b } consisting of characters of GL 1 (F) = F * . Then, Note that π ∨ = π t = π .

Basic properties
An important property of -irreducible representations is the following.
1. By (2), every irreducible subrepresentation of Π also occurs as a quotient of Π. Hence, any irreducible subrepresentation τ of Π such that τ Π is a direct summand of Π. It easily follows that if Π is multiplicity free (i.e., if every element of JH(Π) occurs with multiplicity one), then Π is semisimple, and in particular SA(m, m) is satisfied. 2. In Leclerc's example (12), Π is of length two, and in particular multiplicity free. Hence, in general the condition SA(m, m) is strictly weaker than SSA(m, m) (i.e., the -irreducibility of π ). 3. The first-named author computed JH(Π) for all multisegments m consisting of at most 6 segments. (The computation involves Kazhdan-Lusztig polynomials of the symmetric group S 12 . See [17] for more details.) It turns out that in these cases, Π is multiplicity free if and only if Π has length one or two. An example where this condition fails is in which Π has length 9 with occurring with multiplicity 2 in JH(Π occurring with multiplicity 9 (again, the highest possible) in JH(Π). 4. Unfortunately, in contrast to the multiplicity freeness of Π, we do not have a practical way to completely determine whether Π is semisimple or whether the condition SA(m, m) is satisfied. In fact, at the moment we are unable to refute the condition SA(m, m), or even the semisimplicity of Π, in any single example.
Finally, we mention another simple property, which is a powerful tool for proving -irreducibility ([20, Lemma 2.10]).

Zelevinsky's commuting varieties
We recall Zelevinsky's geometric picture of his classification [16,24,29,30]. Consider a finite-dimensional Irr c -graded C-vector space Up to isomorphism, V is determined by its graded dimension be the group of grading preserving linear automorphisms of V and let → E(V) (resp., be the vector space of (nilpotent) linear transformations A : V → V such that A(V ρ ) ⊆ V→ ρ (resp., A(V ρ ) ⊆ V← ρ ) for all ρ ∈ Irr c . We will use the notational

E(V) is determined by the non-negative integers rk
(which are non-zero for only finitely many i's and ρ's). The orbits are parameterized by the multisegments m such that supp(m) = grdim V. Concretely, where we set x ρ, are also in one-to-one correspondence with the irreducible components of the commuting variety are the canonical projections. Let where we recall that m # was defined in Sect. 2.2. We also write , but the inclusion is strict in general).

The condition G L S(m)
In general, there are infinitely many G(V)-orbits in X(V). Following Geiss-Leclerc-Schröer [10] we make the following definition. By [20,Remark 4.6], the analogous condition for ← C m is equivalent. Therefore, from now on we will exclusively work with → C m . In order to simplify the notation, we will henceforth write C m instead of → C m (and similarly for O m and C • m ).

Linearization
As in [20, §4], it is advantageous to explicate the condition G L S(m) by linearization. More precisely, fix A m ∈ O m and write it in the form (14) for a suitable basis holds if and only if G m admits an open (i.e., dense) orbit in C m , i.e., C m is a prehomogenuous vector space with respect to the action of G m . Passing to the Lie algebra, we can rephrase it by saying that there exists λ ∈ C m such that [g m , λ] = C m where g m is the Lie algebra of G m .
The vector space C m was explicated in [24, Lemme II.4] as follows. Let X m be the set Then, C m admits a basis α i, j = α m i, j , (i, j) ∈ X m given by Similarly, the group G m and its Lie algebra g m are described in [24,Lemme II.5]. Let Y m be the set Then, as a vector space, g m has a basis β i, j = β m i, j , (i, j) ∈ Y m given by We will call g i, j the coordinates of g. Moreover, we have where for convenience we set α r,s = 0 if (r, s) / ∈ X m . Therefore, by passing to the dual map we can explicate the surjectivity of the linear map [·, λ] : g m → C m as follows.
is an open and G m -invariant condition in λ ∈ C m . This condition is also easy to check (by a randomized algorithm) on a computer. (See [20,Remark 4.9] for a hypothetical approach to make this deterministic.) We refer to [20, §4] for examples and further discussion.
The main result of [20] is that Conjecture 3 holds in the special case where In this case, the condition is also related, somewhat surprisingly, to smoothness of Schubert varieties of type A.
It was also verified computationally that Conjecture 3 holds for m consisting of up to 6 segments. This condition is in the spirit of the condition G L S(m) discussed in Sect. 4.2. We conjecture that SG(m, m ) is equivalent to SA(m, m ), at least under the condition (ALO) of Sect. 3.3. In particular, in this case it would give a practical condition for irreducibility of parabolic induction.

General setup
We continue to work with the geometric setup and the notation of the previous section.
Let V and V be two finite-dimensional Irr c -graded vector spaces over C. Consider the direct sumṼ = V ⊕ V and the short exact sequence the resulting P ı -equivariant surjective maps (with U ı acting trivially on the target). (These projections should not be confused with the projections Again, p ı is P ı -equivariant (and in particular, U ı -invariant) and surjective.

Define similarly
by interchanging the roles of V and V . The parabolic subgroups P ı and P ı are opposite with be the diagonal embedding. Its image is

Sum of multisegments
Now let m, m be two multisegments such that supp m = grdim V and supp m = grdim V , and let n = m + m , so that supp n = grdimṼ.
(We do not know whether the above inclusion can be strict.) Similarly, let Let Note that since ⊃ clearly holds while We can now formulate the main geometric conditions.

Proposition 2
The following conditions are equivalent.
Definition 5 We denote the above equivalent conditions by SG(m, m ). (This stands for "subrepresentation+geometric") We denote by IG(m, m ) (for "irre-ducible+geometric") the condition is not satisfied and SG(m , m) is satisfied.
We will give some more equivalent conditions for SG(m, m ) in Sect. 5.4 below and ultimately prove Proposition 2 in Sect. 5.8. In the meantime, we state a simple corollary of Proposition 2. Proof Clearly, IG(m, m ) implies (16a) and its symmetric analog (interchanging m and m ). Conversely, (16b) and the symmetric counterpart of (16a) imply that The corollary follows.

The main conjecture
We now state the new conjecture. Note that the second part of the conjecture would follow from the first one by Corollaries 1 and 2.
More generally, it would be desirable to have a geometric/combinatorial grasp on soc(Z (m)× Z (m )) under (ALO). We hope to get back to this question in the future.
A more ambitious formulation of Conjecture 4 is the following.
Question 1 Are the conditions SA(m, m ) and SG(m, m ) equivalent even without assuming (ALO) ?
An affirmative answer would give that However, it would not directly give an exact irreducibility criterion for π 1 ×π 2 when neither π 1 nor π 2 is -irreducible.

Linearization
Before proving the equivalence of the conditions in Proposition 2 we will first linearize and explicate them in a way similar to what was done for the condition We take a graded basis x ρ,i , ρ ∈ Δ i , i ∈ I (resp., i ∈ I ) for V (resp. V ) for which A m (resp., A m ) has a graded Jordan normal form (14). Thus, A n has a graded Jordan normal form with respect to the union . Thus, C ı n consists of the elements of C n whose X m ,mcoordinates vanish. Define similarly C ı n and let C ı,ı

Proposition 3
We have G n · C ı n = U ı · C ı n and P ı ·C ı,ı n = U ı ·C ı,ı n . Moreover, the following conditions are equivalent.
are linearly independent in the complex vector space Finally, SG(m, m ) is equivalent to the conditions above.
In the rest of the section we will prove Proposition 3.
Remark 3 Assume that Proposition 3 holds. Then, 1. By the same argument as in Corollary 2 (or alternatively, using Remark 4 below), IG(m, m ) is equivalent to the condition for (i, j) ∈ X m and λ ∈ C m .

The linear independence of (17) is an open and
Thus, taking into account (9), in Conjecture 4 we may assume without loss of generality that Z (m) is -irreducible. 5. Condition 3 is the least conceptual but the most practical to check. It can be easily implemented on a computer, at least as a randomized algorithm. Nonetheless, it would be interesting to replace it by a deterministic criterion.

Proof of first part of Proposition 3
Since C ı,ı n is M-invariant and P ı = M U ı , it is clear that P ı ·C ı,ı n = U ı ·C ı,ı n . Also, since U ı P ı is dense in G n (e.g., by considering the Lie algebras) and C ı n is P ı -invariant, we have G n · C ı n = U ı · C ı n . Next, we show the equivalence of conditions 1, 2 and 3 of Proposition 3. We use the following simple general criterion.

Remark 4
Suppose that G is a linear algebraic group with a rational (linear) representation on a finite-dimensional vector space W and let W be a subspace of W . Then, Returning to the case at hand, we identify with C X m,m by considering the X m,m -coordinates. For any μ ∈ C ı n the linear map gives rise to a linear map which depends only on the image (λ, λ ) of μ under the projection C ı n → C m ⊕ C m . Thus, by Remark 4, the conditions 1 and 2 of Proposition 3 are both equivalent to the surjectivity of L μ for some μ ∈ C ı,ı n . Condition 3 is merely an explication of this (or more precisely, of the injectivity of the dual map).

A simple reduction
It remains to show the equivalence of each of the conditions in Proposition 2 with the corresponding condition in Proposition 3, thus completing the proof of the two propositions.
We use the following simple result.

Lemma 2 Let G be a linear algebraic group acting algebraically on quasiaffine varieties X and Y and let p : X → Y be a G-equivariant morphism. Let H be a closed subgroup of G and let W be an H -invariant subvariety of X .
Assume that G acts transitively on Y and fix y 0 ∈ Y . Let G 0 be the stabilizer of y 0 in G, X 0 = p −1 (y 0 ) and W 0 = W ∩ X 0 . Assume that X 0 is irreducible and that W = H · W 0 . Then, G · W = G · W 0 and Proof The first assertion is clear since Conversely, suppose that G · W 0 = X . Then G · W 0 contains an open dense (and without loss of generality, G-invariant) subset U of X . The set is G-invariant and non-empty (and in fact dense in Y ). Since G acts transitively on Y we infer that p −1 (y) ∩ U = ∅ for all y ∈ Y . In particular, X 0 ∩ U is a non-empty open subset of X 0 , hence dense since X 0 is irreducible. A fortiori, where x 0 ∈ X 0 and g ∈ G, then necessarily g ∈ G 0 . Hence, G 0 · W 0 is dense in X 0 as required.

An auxiliary result
We go back to the setup of Sect. 5.2. In order to invoke Lemma 2 we will need an additional result.

Lemma 3
We have the following equalities of spaces.
Proof By the P ı -equivariant of the map → p ı and the surjectivity of the map Hence, the equality (18a) implies (18b) Next, we show that (18b) implies (18c). Indeed, assuming (18b), we have we deduce that Note that the statements with respect to ı are obtained from the original ones by interchanging m and m . Thus, it remains to prove (18a). Since Next, we show that Indeed, recall that On the other hand, for any D ∈ ( Our claim follows.
It remains to show that O n ∩ ( Consider the abelian category whose objects are pairs (U, D) where U is a finite-dimensional Irr c -graded vector space and D ∈ → E(U ); the morphisms between (U, D) and (U , D ) are the grading preserving linear transformations The statement that we need to prove is that if is a short exact sequence and (Ṽ, splits. In fact, this is true for any locally finite C-linear abelian category. Indeed, we have an exact sequence and Comparing dimensions we infer that ı * is onto. Hence, (19) splits.

Conclusion of proof of Propositions 2 and 3
Using Lemmas 2 and 3 we show the equivalence of (16a) and (16b) with the first (resp., second) condition in Proposition 3, thereby completing the proofs of Propositions 2 and 3.
For the first equivalence we apply Lemma 2 to with G = G(Ṽ) and H = P ı . Note that G 0 = G n and the embedding W 0 → X 0 can be identified via ← pṼ with C ı n → C n . By Lemma 3 (with respect to ı ) we have W = H · W 0 and Y ı m ,m = W . Also, by definition, C n = X .
For the second equivalence, we apply Lemma 2 with The condition W = H · W 0 is clear. We have G 0 = U ı and the embedding W 0 → X 0 can be identified using ← pṼ, with C ı,ı n → C ı n . Note that by (15) we have Z m,m = W , and as before, X = Y ı m,m .
It is not difficult to see that SG(m, m ) is also equivalent to each of the following conditions.

The set
We will not give details since we will not use this result.  and Z (m) is -irreducible, then for any m ∈ N(Σ) the irreducibility of Z (m) × Z (m ) is equivalent to IG(m, m ). By remark 3, Conjecture 4 is equivalent to saying that every m such that Z (m) ∈ Irr is good, while Question 1 asks whether in fact every m ∈ N(Σ) is good.
Recall that a multisegment m is called a (strict) ladder if it can be written as Δ 1 + · · · + Δ k where Δ i+1 ≺ Δ i for all i = 1, . . . , k − 1. This class is invariant under m → m # . The corresponding representations were studied in [18,19]. In particular, G L S(m) is satisfied and Z (m) is -irreducible for any ladder m.
The following result provides plenty of examples of good multisegments, in support of Conjecture 4.  Indeed, the first part follows from Tadić classification [27] (see also [19]) which implies that if Z (m) is unitarizable, then we can write m as a sum of multisegments The second part follows from [21, Proposition 2.7]. Finally, if Z (m) is generic (resp., unramified), then every segment is m (resp., m # ) is a singleton.
Another case which will be useful later on is the following.
Example 2 Suppose that m = Δ 1 +· · ·+Δ k and there exists ρ ∈ Irr c such that b(Δ i ) ∈ {ρ, → ρ } for all i. Then, we can write m = m 1 +· · ·+m l such that each m i is a ladder (consisting of either one or two segments) and Z (m i ) × Z (m j ) is irreducible for all i, j. (See Sect. 8.7 below.) Thus, m is good.
We will prove the theorem (along with all other results stated in this section) in Sect. 8 below.

More precise statements
The proof of Theorem 1 depends on several compatibility properties of the conditions SA(m, m ) and SG(m, m ) which are interesting in their own right. We formulate them in the following propositions, which provide additional attestation for Conjecture 4 as well as for an affirmative answer to Question 1.
Recall the notation max m and m − from Sect. 2.2. The first part of Theorem 1 will follow from Proposition 4 and the following.
. Assume that I = I 1 ∪ I 2 and I = I 1 ∪ I 2 (disjoint unions) and that for any i ∈ I 1 ∪ I 1 and j ∈ I 2 ∪ I 2 for any i ∈ I 1 ∪ I 1 and j ∈ I 2 ∪ I 2 we have Let m r = i∈I r Δ i and m r = i ∈I r Δ i , r = 1, 2. Then,

G L S(m) implies G L S(m 1 ) and G L S(m 2 ).
Extending the partial order on Irr c lexicographically to Σ (see Sect. 7

G L S(m) implies G L S( i∈I
The second part of Theorem 1 follows from the following more precise statement.

Suppose that Z (m) and Z (m ) are -irreducible and Z (m) × Z (m ) is irreducible. Then
SA(m + m , n) ⇐⇒ SA(m, n) and SA(m , n).
By passing to the contragredient, similar statements hold for the relations obtained from SA or SG by switching the arguments.
Note that the statements of the proposition are not completely symmetric in SA and SG. It would be interesting to show the missing representationtheoretic counterparts, namely SA(m, m ) and SA(m + m , n) ⇒ SA(m, n), SA(m, m ) and SA(m, n) ⇒ SA(m, m + n) at least when Z (m), Z (m ) and Z (n) are -irreducible.

Further evidence
The following compatibility result is motivated by (13). For the last consistency check that we will state here, we fix ρ ∈ Irr c . For π ∈ Irr we write ρ π if there does not exist π ∈ Irr such that π → ρ × π . (A more general notation will be introduced in Sect. 7.2 below.) For any π = Z (m) ∈ Irr there exist a unique integer m ≥ 0 and a unique π = Z (m ) ∈ Irr such that π → m ρ × · · · × ρ ×π and ρ π . We denote ρ m = m . On the level of the Grothendieck group, this gives rise to a ring homomorphism

For instance, for any segment
By Frobenius reciprocity, for any π, π ∈ M we have a canonical functorial surjection [5] p π,π : J (π × π ) → π ⊗ π . For any s ∈ N(Irr c ) let M s be the Serre subcategory of M consisting of representations all of whose irreducible subquotients have supercuspidal support s. We have and M s × M s ⊂ M s+s for any s, s ∈ N(Irr c ). For any π ∈ M and s ∈ N(Irr c ) we denote by π s the s-component of π with the respect to the decomposition (22). Thus, Similarly, for any Π ∈ M ⊗ M and s, s ∈ N(Irr c ) we denote by Π s⊗s the s ⊗ s -component of Π with the respect to the decomposition

Remark 7
Let π ∈ M, s ∈ N(Irr c ) and τ ∈ M s . Any morphism p : π → τ factor through a morphism π s → τ . Suppose that p is surjective. Then, the restriction p s of p to π s is surjective. Moreover, the following conditions are equivalent.
In this case, we say that p is a component map, or that (somewhat informally) τ "is" the s-component of π . Similar terminology will apply for M ⊗ M and s ⊗ s with s, s ∈ N(Irr c ).

Separated representations
The following technical definition will be useful.

Definition 6
Suppose that π ∈ M s and π ∈ M s for some s, s ∈ N(Irr c ). We write π π and say that π is left-separated from π if p π,π is a component map, i.e., if the following equivalent conditions are satisfied.
As the notation suggests, the relation is not symmetric.
The following easy property will be used repeatedly.
Proof Suppose that we have a short exact sequence Then, we have a commutative diagram where the rows are exact and the vertical morphisms are surjective. It follows that π π if and only if σ π and τ π . A similar statement holds for a short exact sequence The lemma immediately follows.
We will use two additional properties of the relation .

A simple criterion for separation
Recall that we fixed a total order ≤ on Irr c subject only to the condition that ρ < → ρ for all ρ ∈ Irr c . We extend this lexicographically to a total order on Σ (also denoted by ≤), that is, Δ ≤ Δ iff either e(Δ) < e(Δ ) or e(Δ) = e(Δ ) If m = i∈I Δ i ∈ N(Σ) and Δ ∈ Σ, then we write m ≥Δ = i∈I :Δ i ≥Δ Δ i and similarly for m <Δ . Proof We show that z(m) z(m ), the other part being similar. Suppose on the contrary that this is not the case. Then, by the geometric lemma of , there exists for each i ∈ I ∪ I a partition Δ i = A i ∪ B i such that ρ 1 < ρ 2 whenever ρ 1 ∈ A i and ρ 2 ∈ B i , i ∈ I ∪ I . In particular, and let i ∈ I be such that ρ ∈ B i . Then, e(Δ i ) ∈ B i and therefore there exists i ∈ I such that e(Δ i ) ∈ A i . Thus, b(Δ i ) ∈ A i and we infer that b(Δ i ) ≥ ρ by the minimality of ρ. Hence, (3)) and therefore ← Δ i ≺ Δ i , in contradiction to our assumption.

Proofs
In this section we prove the statements of Sect. 6.
For the properties pertaining to the geometric condition SG(m, m ) we will mostly use the last and most tangible criterion of Proposition 3. It is likely, however, that the proofs can be made more conceptual.
When writing a multisegment as m = i∈I Δ i it will sometimes be convenient to allow Δ i = ∅ for some indices. These inconsequential indices will not have any effect on m or on the objects pertaining to it (such as X m , Y m of Sect. 4.3). We will use this convention throughout.
To prove the third part, let λ ∈ C m and λ ∈ C m . For r = 1, 2, letλ ∈ C m r be such thatλ i, j = λ i, j for all (i, j) ∈ X m r . Defineλ ∈ C m r similarly. Clearly, for any (i, j) ∈ X m r ,m r , the Y m r ,m r -coordinates of v m,m i, j (λ, λ ) coincide with those of v m r ,m r i, j (λ,λ ). On the other hand, the assumptions (20a) and (20b) imply that the coordinates outside Y m r ,m r vanish. This clearly implies the third part of Proposition 5.
The last part of Proposition 5 is proved similarly. Corollary 4 is now an immediate consequence.

Proof of first part of Proposition 4
For any 0 = m = i∈I Δ i ∈ N(Σ) we write m = m mx + m nmx where m mx = i∈I :e(Δ i )=max m Δ i and m nmx = i∈I :e(Δ i )<max m Δ i . For consistency we also write 0 mx = 0 nmx = 0. This follows from (10) and Lemmas 5 and 7.
Proof The first equality of (28) follows from Lemma 8 while the last one follows from Lemma 6, Corollary 6 and the condition on π . It remains to show that any irreducible subrepresentation π of π × L(m nmx ) × L(m mx ) is contained in π × L(m). Assume on the contrary that this is not the case. Then π → π × L(m ) for some L(m ) ≤ L(m nmx ) × L(m mx ) with m mx = m mx (by the second part of Lemma 8). On the other hand, by the last equality of (28) we have π → τ × L(m mx ) for some irreducible τ → π × L(m nmx ) and hence by Lemma 8 we have m mx = m mx where we write π = L(m ). Now, π → π × L(m nmx ) × L(m mx ). Hence, π → τ × L(m mx ) for some irreducible subquotient τ of π × L(m nmx ). It follows from Lemma 8 that m mx = m mx = m mx . We get a contradiction.
The last part follows from (28) and Lemma 8. Indeed, if max(m−Δ) = max m, i.e., if m mx = Δ then this is just Lemma 9; otherwise this follows from Lemma 9 by applying it to both m and m − Δ (with m and m − Δ respectively).
In fact, the following slightly more precise statement holds (although we shall not use it here).
The corollary follows.
We can now prove the first part of Proposition 4, which is a strengthening of [19,Lemma 4.16].
It remains therefore to show that SA(m, m ) implies (m + m ) − = m + m − . The argument is essentially in [19]. For completeness we recall it. Write m = i∈I Δ i , m = i ∈I Δ i and let i 1 , . . . , i m be leading indices of m . Assume on the contrary that (m+m ) − = m+m − . Then, there exist indices l ≤ m and By Corollary 4 we may replace m and m by m ≥Δ l and m ≥Δ l respectively since neither the condition SA(m, m ) nor the assumption (m + m ) − = m + m − is affected by this change. Thus, we may assume without loss of generality that m = m ≥Δ l and l = m . By Lemma 9 (applied to m # ), the condition SA(m, m ) implies that However, the right-hand side admits a segment which contains e(Δ l ) in its support while the left-hand side does not. We obtain a contradiction. Proof We argue as in [24]. We prove by induction on l = 0, . . . , m − 1 that there exists λ ∈ A such λ i,i j = δ i,i j+1 for all j = 1, . . . , l and i ∈ I such that e(Δ i ) = e(Δ i j+1 ). The base of the induction (l = 0) is trivial. For the induction step, we assume that 0 < l < m and the statement holds for l − 1. In addition, by openness, we may assume that λ i l+1 ,i l = 0. Take g ∈ G m whose coordinates are given by (g is invertible since λ i l+1 ,i l = 0.) Letλ = g −1 λg ∈ A. We show thatλ satisfies the required conditions for l. Suppose first that j < l. By assumption, for any ρ ∈ Δ i j we have λx ρ,i j = x← ρ ,i j+1 + ξ where the coordinates of ξ with respect to x← ρ ,i vanish if e(Δ i ) = e(Δ i j+1 ). The same condition holds for and hence,λ i,i j = δ i,i j+1 for all i such that e(Δ r ) = e(Δ i j+1 ). Also, we may write It follows thatλ i,i l = δ i,i l+1 for all i such that e(Δ i ) = e(Δ i l+1 ). This concludes the induction step.
Similarly, we have the following. We show by descending induction on l = m, . . . , 1 that there exists λ ∈ B such that λ i j ,i = 0 for all j > l and i such that e(Δ i j ) = e(Δ i ). Once again, the base of the induction (l = m) is trivial.
For the induction step we may assume that Δ i l−1 is not a singleton (or equivalently, (i l , i l−1 ) ∈ X m − ) since otherwise, (i l , i) / ∈ X m − for all i such that e(Δ i j ) = e(Δ i ). We may also assume that λ ∈ B satisfies λ i l ,i l−1 = 0. Consider the set Note that it follows from the definition of i l that Then, for any ρ ∈ Δ i , i ∈ I l the coefficients of with respect to x← ρ , j are 0 for all j such that Δ j ≥ Δ i l . Hence, the coefficient It follows from the induction hypothesis thatλ i j ,i = λ i j ,i = 0 for all i ∈ I j . This completes the induction step.
Recall the sets X m,m and Y m,m defined in Sect. 5.4. Assume first that SG(m, m ) holds. Thus, v m,m i, j (λ, λ ), (i, j) ∈ X m,m are linearly independent in C Y m,m for some (λ, λ ) ∈ C m × C m (see (17)). Let

Lemma 12 Suppose that
)}, Note that X m \X m − =X m ∪X m . By Lemma 10 (applied to m ) we can assume without loss of generality that λ i, j = 0 for all (i, j) ∈X m .
Let pr : , and let f = f m,m :Ỹ m,m →X m,m be the bijection defined in Lemma 12. By Lemma 11 we may assume without loss of generality thatλ i, j = 0 if (i, j) ∈ X m − \X m . We may also assume thatλ i j +1 ,i j = 0 for all j = 1, . . . , m − 1 such that Δ i j is not a singleton. Define λ ∈ C m by As before, To conclude SG(m, m ) it suffices to show that v m,m i, j (λ, λ ), (i, j) ∈X m,m are linearly independent. Indeed, for any To complete the proof of Proposition 4 (the very last part), it remains to invoke the penultimate part of Corollary 4.

Proof of first part of Theorem 1
We show by induction on k that the conditions SA(m, m ) and SG(m, m ) are equivalent for any multisegment m = i∈I Δ i . (The equivalence of the symmetric conditions SA(m , m) and SG(m , m) would then follow from (9) and Remark 3 part 4.) The case k = 0 is trivial. Assume that k > 0 and the statement holds for k −1. By Proposition 4 we may assume that max m ≤ max m. In this case, by [19, Proposition 6.1] SA(m, m ) is equivalent to SA(m − Δ 1 , m <Δ 1 ). To complete the induction step we show that the conditions SG(m, m ) and SG(m − Δ 1 , m <Δ 1 ) are also equivalent.
Note that Suppose that λ i, j =λ i, j for all (i, j) ∈ X m−Δ 1 and λ i, j =λ i, j for all (i, j) ∈ X m <Δ 1 . Then, it is easy to check that for all (i, j) ∈ X m−Δ 1 ,m <Δ m ). This completes the induction step.

Remark 8
In [19,Proposition 6.15] we also proved that if at least one of m and m is a ladder then SG(m, m ) is equivalent to the condition that there exists an injective map f : In general, the latter condition is strictly weaker than SG(m, m ).

Proof of Proposition 6 and second part of Theorem 1
and Z (m+m +n) Z (m)× Z (m +n), we infer that necessarily π = Z (m +n). Hence, SA(m, m + n). Now suppose that SG(m + m , n) holds. As before, we use the criterion 3 of Proposition 3. Let λ ∈ C m+m and λ ∈ C n be such that v m+m ,n i, j (λ, λ ), (i, j) ∈ X m+m ,n are linearly independent in C Y m+m ,n . Denote by λ m the element of C m whose coordinates coincide with the X m -coordinates of λ. Similarly for λ m . If SG(m, m ) holds, then by Proposition 3, we may assume that λ i, j = 0 for all (i, j) ∈ X m,m . (We may still assume that (λ m , λ m ) ∈ C m × C m is generic.) In this case, the Y m ,n -coordinates of v m+m ,n For the third part, write X m +n = X m ∪ X n ∪ X m ,n ∪ X n,m and take λ ∈ C m +n whose coordinates with respect to X m ,n ∪ X n,m vanish.
The fourth part is an immediate consequence.
To prove the last part of Proposition 6, recall that SA(m + m , n) if and only if the image of the intertwining operator R Z (n),Z (m+m ) is Z (m + m + n). On the other hand, by assumption Z (m + m ) = Z (m) × Z (m ) and therefore The required equivalence follows from the fact that Z (m + m + n) Z (n) × Z (m + m ), JH(Z (n + m) × Z (m )) and JH(Z (n + m ) × Z (m)) and that the images of R Z (n),Z (m+m ) , R Z (n),Z (m) and R Z (n),Z (m ) are irreducible.
This finishes the proof of Proposition 6. The second part of Theorem 1 now follows by induction on k.

Proof of Proposition 7
The first part is a special case of (13). We prove the second part. Observe that Let λ m and λ m be as in Sect. 8.5. Then, for any ∈ X m and 0 otherwise. 3. If (i, j) ∈ X m,m then the Y m,m -coordinates of v n i, j (λ) are v m,m i, j (λ m , λ m ). It follows from the assumptions G L S(m ) and SG(m, m ) that in order to show G L S(n), it is enough to know that v n,m i, j (λ, λ m ), (i, j) ∈ X n,m are linearly independent for generic λ ∈ C m ,m . By the condition SG(n, m), v n,m i, j (λ, λ ), (i, j) ∈ X n,m are linearly independent for generic (λ, λ ) ∈ C n × C m . Since this condition in λ is invariant under the action of G n , it follows from Proposition 3 and the condition SG(m, m ) that v n,m i, j (λ, λ ), (i, j) ∈ X n,m are linearly independent for generic (λ, λ ) ∈ C ı n ×C m . Since this condition in λ is invariant under the action of G m , the condition G L S(m) (which says that C m admits an open G m -orbit) guarantees that we can take λ = λ m , as required.

Proof of Proposition 8
For this subsection fix ρ ∈ Irr c .
We now recall some results from [13,23] (see also [19,20]). Let m = i∈I Δ i ∈ N(Σ). We say that two subsets A and B of I are equivalent and write We think of R as a partially defined bijection and we write R(i) = j and Given two ρ-matchings R 1 , R 2 we say that R 2 dominates R 1 if one of the following conditions holds.
The transitive closure of domination is a partial order on the set of ρ-matchings, which we denote by ≤. We say that two ρ-matchings R 1 and R 2 are equivalent if Maximal ρ-matchings are not unique up to equivalence. (For instance, we could take m = Δ 1 + Δ 2 + Δ 3 + Δ 4 such that X (Technically, they are only defined up to equivalence, but this will not matter in what follows.) Moreover, A best ρ-matching exists and is unique up to equivalence. Moreover, a maximal ρ-matching R is best if and only if the product of the representations It is useful to introduce another notion which is weaker than maximality.

Definition 7
We say that a ρ-matching R for m is saturated if X m ∩ (A(R) × B(R)) = ∅ and for every (i, j) ∈ R and i ∈ A(R) such that (i , j) ∈ X m we have Δ i ≤ Δ i .

Lemma 13
Let R be a saturated ρ-matching R for m. Then, for any R ≥ R, R is saturated and A(R) = A(R ). Hence, A(R) ∼ A ρ m . Proof It is enough to show it when R dominates R. By the saturation of R it is clear that R is obtained from R by replacing a certain pair (i, j) ∈ R by (i, j ) ∈ X m where j ∈ B(R) and Δ j < Δ j . Thus, A(R) = A(R ) and B(R ) = B(R)∪{ j}\{ j }. Let us show that R is saturated. First, if (i , j) ∈ X m with i ∈ A(R) then Δ i ≤ Δ i by saturation and hence (i , j ) ∈ X m (since (i, j ) ∈ X m ). Thus, X m ∩(A(R )× B(R )) = ∅. On the other hand, we cannot have i ∈ A(R) with (i , j ) ∈ X m since R is saturated. Therefore, the second condition for saturation for R follows from the saturation of R.
In both cases i∈A ρ m+m Δ i < i∈A ρ m Δ i . Conversely, suppose that f is onto. Then, R ∪ R is saturated (with respect to m + m ). It follows from Lemma 13 that A ρ m+m ∼ A ρ m as required. The following result follows from Lemma 5.

Lemma 16
Let R be a ρ-matching for m. Then, any non-empty open G minvariant subset S of C m contains an element λ such that λ i,R( j) = δ i, j for all Proof Let { j 1 , . . . , j r } be the domain of R with Δ j 1 ≥ · · · ≥ Δ j r . We show by induction on l that we can find λ ∈ S such that λ i,R( j m ) = δ i, j m for all m = 1, . . . , l and all i ∈ Y ρ m such that (i, j m ) ∈ Y m . The case l = 0 is trivial. For the induction step, let 0 < l ≤ r and assume that the statement holds for l − 1. By openness, we may assume in addition that λ satisfies λ j l ,R( j l ) = 0. Define g ∈ G m by Note that g is invertible because λ j l ,R( j l ) = 0. Letλ = g −1 λg. Then, gx→ ρ ,R( j l ) = x→ ρ ,R( j l ) and It follows thatλ i,R( j l ) = δ i, j l for all i ∈ Y ρ m such that (i, j l ) ∈ Y m . Now let m < l. Then, gx→ ρ ,R( j m ) = x→ ρ ,R( j m ) and by induction hypothesis λx→ ρ ,R( j m ) = x ρ, j m + ξ where the coordinate of ξ with respect to x ρ,i is zero unless i / ∈ Y ρ m or Δ i > Δ j m . Since Δ j l ≤ Δ j m it follows that λx→ ρ ,R( j m ) = λx→ ρ ,R( j m ) = x ρ, j m + ξ.
This completes the induction hypothesis.
By passing to the contragredient we get Suppose that SG(m, m ). Let (λ, λ ) ∈ C m × C m be such that v m,m i, j (λ, λ ), (i, j) ∈ X m,m are linearly independent in C Y m,m . Note that the Y ρ m,mcoordinates of v m,m i, j (λ, λ ), (i, j) ∈ X m,m are independent of theX mcoordinates of λ. By Lemma 16 (for R maximal) we may assume that λ i, j = 0 for all (i, j) ∈X Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.