joint with Alberto Mínguez

Transcription

1 joint with Alberto Mínguez

2 We consider (complex, smooth, finite length) representations of GL n (F ) where F is a non-archimedean local field. We write Irr = n 0 Irr GL n (F ) (equivalence classes of irreducible representations). JH(π) is the Jordan-Hölder sequence of π (a multiset of elements of Irr). We denote by π 1 π 2 the representation parabolically induced from π 1 π 2 (normalized induction). soc(π) is the socle of π, i.e., the sum of the irreducible subrepresentations of π (= the maximal semisimple subrepresentation of π). (If π 0 then soc(π) 0.) We say that π is SI if soc(π) is irreducible and occurs with multiplicity one in JH(π).

3 For any two representations π i of GL ni (F ), i = 1, 2 we have the standard intertwining operator M π1,π 2 (s) : π 1 s π 2 π 2 π 1 s. The integral defining M π1,π 2 (s) converges for R(s) 0 and defines a rational function in q s. Let r π1,π 2 = ord s=0 M π1,π 2 (s) 0 and R π1,π 2 = lim s 0 s rπ 1,π 2 Mπ1,π 2 (s) : π 1 π 2 π 2 π 1 (defined and non-zero).

4 Theorem (Kang, Kashiwara, Kim, Oh, 2014) The following conditions on a representation π of GL n (F ) are equivalent. 1. π π is irreducible. 2. End GL2n (F )(π π) = C. 3. R π,π is a scalar. 4. π π is SI. Moreover, under these conditions, for any irreducible σ of GL m (F ), π σ and σ π are SI; soc(π σ) = Im R σ,π, soc(σ π) = Im R π,σ. The proof is elementary and is based on the following Lemma: Let π i be representations of GL ni (F ), i = 1, 2, 3. σ π 1 π 2 τ π 2 π 3 Assume σ π 3 π 1 τ (submodules of π 1 π 2 π 3 ). Then ω π 2 such that σ π 1 ω and ω π 3 τ.

5 The class of representations π such that π π is irreducible has some nice properties (also proved by KKKO). For instance, if π i π i is irreducible, i = 1, 2 then π := π 1 π 2 is irreducible if and only if R π1,π 2 is an isomorphism, in which case π π is irreducible.

6 There are examples of irreducible π s such that π π is reducible. The first one was constructed by Leclerc (2003), giving a counterexample to a conjecture by A. Berenstein Zelevinsky (1993). Let σ = soc( det 1 2 GL2 det 1 2 GL 2 ) Irr GL 4. (A Speh representation.) Then σ det σ det 1 (a representation of GL 8 ) has length 3. Its socle is (another Speh representation) π 1 = soc( det 3 2 GL2 det 1 2 GL2 det 1 2 GL 2 det 3 2 GL 2 ). Its cosocle is (yet another Speh representation) π 2 = soc( det 1 2 GL4 det 1 2 GL 4 ). Let π = Ker R σ det,σ det 1/ Im R σ det 1,σ det Irr GL 8 be the remaining element of JH(σ det σ det 1 ). Then π π is semisimple of length 2. (One of its constituents is π 1 π 2.)

7 Questions Is R π,π an isomorphism for any π Irr? Is π π semisimple for any π Irr?

8 Classification of Irr (Bernstein Zelevinsky) Let ρ Irr be supercuspidal. ρ = ρ, ρ = ρ 1. If ρ Irr is supercuspidal then ρ ρ is reducible if and only if ρ is either ρ or ρ. A segment is a non-empty set of the form = {ρ 1,..., ρ k } where ρ i Irr is supercuspidal and ρ i+1 = ρ i. In this case ρ 1 ρ k is SI. Let Z( ) = soc(ρ 1 ρ k ). We have Z( ) = Z( ) where = {ρ k,..., ρ 1 }. Notation: b( ) = ρ 1, e( ) = ρ k, = { ρ 1,..., ρ k }, = { ρ 1,..., ρ k }. 1 and 2 are linked if 1 2 is a segment which properly contains each of 1 and 2. In this case, either b( 2 ) 1 (and we write 1 2 ) or the symmetric condition. If 1 and 2 are linked, we can form the descendant pair of segments ( 1 2, 1 2 ), possibly with 1 2 =.

9 The classification theorem A multisegment is a formal sum m = N of segments. Assume that 1,..., N is a sequence of segments such that i j for all i < j. Then ζ(m) := Z( 1 ) Z( N ) (standard module) is SI and depends only on m = N. The map m Z(m) := soc(ζ(m)) defines a bijection between multisegments and Irr. Bernstein Zelevinsky (1977), Zelevinsky (1980).

10 Some properties of Z(m) Z(m) = ζ(m) ζ(m) is irreducible m is pairwise unlinked, i.e., no two i s in m are linked. Write n m if n is obtained from m by replacing a pair of linked segments in m by their descendants. The transitive closure of this relation is denoted by =. (In particular, m = m.) The elements of JH(ζ(m)) are Z(n), n = m. The multiplicities are more subtle. (More about that later.) There is a unique pairwise unlinked multisegment m ul = m. Z(m ul ) occurs with multiplicity one in JH(ζ(m)). Z(m) = Z(m ) where ( N ) := N. Using 1 instead of we get the Langlands classification m L(m). Z(m + n) occurs with multiplicity one in JH(Z(m) Z(n)). In particular, if Z(m) Z(n) is irreducible then it equals Z(m + n).

11 From now on we fix ρ Irr supercuspidal and only consider segments and multisegments supported in the cuspidal line of ρ Z ρ = {ρ n : n Z}. For integers a b we write [a, b] = [a, b] ρ = {ρ a,..., ρ b }. Since ρ is fixed throughout, we will suppress it from the notation. (For all practical purposes ρ is the trivial character of GL 1.) In fact ρ is immaterial the Hecke algebra pertaining to the m {}}{ Bernstein component of ρ ρ is essentially independent of ρ (i.e., it is the usual Iwahori Hecke algebra for S m ) (either type theory or Heiermann, 2011). It will be convenient to set Z([a, a 1]) = 1 (the irreducible one-dimensional representation of GL 0 ) and Z([a, b]) = 0 (the trivial, zero-dimensional representation) if b < a 1.

12 Let m = [a 1, b 1 ] + + [a N, b N ] be a multisegment. 1. m is called regular if #{a 1,..., a N } = #{b 1,..., b N } = N. (We do not exclude a i = b j for some i, j.) 2. Note that if m is regular then n is regular for any n = m. 3. We say that m is almost pairwise unlinked (APU) if m ul m. 4. The complexity of a multisegment m (denoted c(m)) is the maximal integer l 0 for which there exists a chain of multisegments (m ul =)m l... m 1 m. 5. The depth of m (denoted d(m)) is the number of APU multisegments = m. 6. If m is regular then d(m) c(m). We say that m is balanced if d(m) = c(m). For instance, d(m) = 0 c(m) = 0 m is pairwise unlinked. Caution: an APU multisegment can have depth (as well as complexity) > 1.

13 Example: m = [ 1, 4] + [ 2, 3] + [ 3, 2] + [ 4, 1] (Speh representation). Chain of maximal length: m ul =[ 4, 4] + [ 3, 3] + [ 2, 2] + [ 1, 1] [ 4, 3] + [ 3, 4] + [ 2, 2] + [ 1, 1] [ 4, 2] + [ 3, 4] + [ 2, 3] + [ 1, 1] [ 4, 1] + [ 3, 4] + [ 2, 3] + [ 1, 2] [ 4, 1] + [ 3, 3] + [ 2, 4] + [ 1, 2] [ 4, 1] + [ 3, 2] + [ 2, 4] + [ 1, 3] m. The APU multisegments = m: [ 4, 3] + [ 3, 4] + [ 2, 2] + [ 1, 1], [ 4, 2] + [ 3, 3] + [ 2, 4] + [ 1, 1], [ 4, 1] + [ 3, 3] + [ 2, 2] + [ 1, 4], [ 4, 4] + [ 3, 2] + [ 2, 3] + [ 1, 1], [ 4, 4] + [ 3, 1] + [ 2, 2] + [ 1, 3], [ 4, 4] + [ 3, 3] + [ 2, 1] + [ 1, 2].

14 Thus, d(m) = c(m) = 6 and m is balanced. More generally any m = [a 1, b 1 ] + + [a N, b N ] with a 1 > > a N, b 1 > > b N (ladder case) is balanced. (The complexity and depth are ( N 2) if a1 b N + 1.)

15 Leclerc s Example: m = [1, 2] + [ 1, 1] + [0, 0] + [ 2, 1] Chain of maximal length: m ul = [ 2, 2] + [ 1, 1] [ 1, 2] + [ 2, 1] [ 1, 2] + [0, 1] + [ 2, 1] [ 1, 2] + [1, 1] + [0, 0] + [ 2, 1] m = c(m) = 4. The APU multisegments = m are: [ 2, 2] + [1, 1] + [ 1, 0], [1, 2] + [ 1, 1] + [ 2, 0], [ 2, 2] + [0, 1] + [ 1, 1], [0, 2] + [ 1, 1] + [ 2, 1], [ 1, 2] + [ 2, 1] Therefore d(m) = 5. Thus, m is not balanced.

16 Thm ( -Mínguez) Suppose that m is regular and balanced. Then Z(m) Z(m) is irreducible. Remark We expect that in the regular case, the converse also holds. In the non-regular case we do not currently have a good guess for the characterization of the irreducibility of Z(m) Z(m). The theorem is proved by induction. For the induction step we use the following Lemma Let π Irr. Assume π 1, π 2 Irr such that: π π1 π 2, πi π i is irreducible, i = 1, 2, π π 1 is irreducible. Then π π is irreducible. We take π 1 to be a ladder representation. In order to use the lemma we need practical irreducibility criteria.

17 Ladder representations Ladder multisegments are those of the form m = N i = [a i, b i ] where a 1 > > a N and b 1 > > b N. A particularly important example: a i+1 = a i 1, b i+1 = b i 1. (Speh representations) Z(m) Z(m) is irreducible. Determinantal formula (in the Grothendieck group): Z(m) = w S N sgn w Z([a 1, b w(1) ]) Z([a N, b w(n) ]) = det(z([a i, b j ])). Any Jacquet module of a ladder representation is a direct sum of ladder representations (explicitly described) with distinct cuspidal support.

18 Example The Jacquet module is the tensor product of the four ladders composed of the four different colors. Roughly speaking, ladder representations behave very much like square-integrable (or segment) representations. Theorem (Max Gurevich 2016) The product of two ladder representations is multiplicity free.

19 A geometric interpretation Start with a 1 < < a N, b 1 > > b N. To avoid trivialities, assume that a i b N+1 i + 1 for all i. For any σ S N consider m σ = N [a σ 1 (i), b i ]. i=1 (Any regular multisegment is of this form.) It is possible that m σ will be void (i.e., if b i < a σ 1 (i) 1 for some i). However, m w0 is not void. (It is a ladder.) Let σ 0 S N be the (unique) permutation such that m σ0 is pairwise unlinked. (If a N b N + 1 then σ 0 = id. This is the saturated case.) Then m σ is not void if and only if σ σ 0 in the Bruhat order. Example: N = 4, a i = b i = i 3. Then σ 0 = (1243), m σ0 = [ 2, 2] + [ 1, 1] + [1, 0] + [0, 1].

20 Let τ = (τ 1... τ l ) S l and σ = (σ 1... σ n ) S n, n l. We say that τ occurs in σ if there exist indices i 1 < < i l such that the order type of σ i1,..., σ il coincides with τ, that is σ ij < σ ik if and only if τ j < τ k. In other words, the permutation matrix of τ occurs as a l l-minor in the permutation matrix of σ. For instance, τ = (4231) occurs in σ = ( ). If τ does not occur in σ, we say that σ avoids τ. The possible σ 0 s obtained as above are exactly the 213-avoiding permutations. Their number is the Catalan number C N = 1 ( ) 2N = N + 1 N ( 2N N ) ( 2N N + 1 ).

21 For any σ S N let X σ be the Schubert variety corresponding to σ, i.e., the closure of the Schubert cell Y σ = B N (C)\B N (C)T σ B N (C) in the flag variety B N (C)\ GL N (C). Thus, X id is a point and X w0 = B N (C)\ GL N (C). In general X σ is projective, but not smooth. X τ X σ τ σ in the Bruhat order. If moreover, σ 0 τ this is equivalent to m τ = m σ. Thus, for any σ σ 0, c(m σ ) = l(σ) l(σ 0 ) while d(m σ ) = #{i < j : σ 0 (i) > σ 0 (j) and σ 0 τ i,j σ} where τ i,j is the transposition i j. The following conditions are equivalent for any σ 0 σ. 1. Y σ0 is contained in the smooth locus of X σ. 2. #{i < j : σ 0 (i) > σ 0 (j) and σ 0 τ i,j σ} = l(σ) l(σ 0 ). 3. P σ0,σ = 1 (Kazhdan Lusztig polynomial). 4. P σ,σ = 1 for all σ 0 σ σ. (Lakshmibai Seshadri (1984), Deodhar (1985))

22 Thus, we can rephrase our theorem as follows: If σ σ 0 and X σ is smooth at Y σ0 then Z(m σ ) Z(m σ ) is irreducible. Alternatively, we can write the condition as the determinantal formula Z(m σ ) = sgn σσ ζ(m σ ). σ 0 σ σ Note that for Leclerc s example N = 4, σ = (4231), σ 0 = (1243). In this case P σ0,σ(q) = 1 + q.

23 Relation to category O (Arakawa Suzuki (1998)) (See also Henderson (2007), Barbasch Ciubotaru (2015)) Let λ = (λ 1,..., λ N ) Z N with λ 1 λ N. There is an exact functor F λ from category O with respect to gl N to representations supported in the cuspidal line of ρ. It satisfies F λ (M(µ)) = ζ(m µ,λ ) { Z(m µ,λ ) if µ i µ i+1 whenever λ i = λ i+1, and F λ (L(µ)) = 0 otherwise, where m µ,λ = N i=1 [µ i, λ i ]. We may apply the Kazhdan Lusztig conjecture (proved by Beilinson Bernstein, Brylinski Kashiwara (1981)) expressing the change of basis matrix between the irreducible modules L(µ) and the Verma modules M(µ) in category O in terms of the Kazhdan Lusztig polynomials. Then we can apply F λ. Let µ = (µ 1,..., µ N ) Z N with µ 1 µ N. Note that m wµ,λ m w µ,λ in the Zelevinksy order (i.e., Z(m w µ,λ) occurs in ζ(m wµ,λ )) if and only if w w in the Bruhat order of S N.

24 Let S λ = stabilizer of λ in S N. We get that for any µ = (µ 1,..., µ N ) Z N with µ 1 µ N and w S N we have ζ(m wµ,λ ) = w [S λ \S N /S µ] P w,w (1) Z(m w µ,λ) where [S λ \S N /S µ ] = {w S N of maximal length in S λ w S µ } and P w,w are the Kazhdan Lusztig polynomials wrt S N. Equivalently, for w [S λ \S N /S µ ] we have Z(m wµ,λ ) = w S N sgn w w P w w 0,ww 0 (1) ζ(m w µ,λ) where w 0 = longest element in S N. Ladder representations correspond to S λ = S µ = 1 and w = 1. The determinantal formula reflects the fact that P w,w 0 = 1 for all w S N. In fact, one gets a resolution of a ladder by standard modules (corresponding to the BGG resolution).

25 The irreducibility of π π when π is a ladder is translated to the following identities. Let H be the parabolic subgroup of S 2N H = {w S 2N : {w(2i 1), w(2i)} = {2i 1, 2i} i} S N 2. H\S 2N /H {M Mat N (Z 0 ) : N M i,j = i=1 N M k,l = 2 j, k} l=1 = {T σ1 + T σ2 : σ 1, σ 2 S N } where T σ = permutation matrix of σ. w σ 1 σ 1 2 defines a map H\S 2N /H conj.cls.(s N ), w [w]. Then x HwH sgn x P x, w0 (1) = sgn[w] 2 r w H\S 2N /H where r is the number of cycles of size > 1 in [w] and for any σ S N, σ S 2N is the doubled permutation σ(2i 1) = 2σ(i) 1, σ(2i) = 2σ(i).

26 Remark The map w [w] factors through a bijection K\S 2N /K conj.cls.(s N ) where K = N S2N (H) = H S N = C S2N (2i 1 2i, 2i 2i 1). I am not aware of an independent proof of the relation sgn x P x, w0 (1) = sgn[w] 2 r x HwH even for w = id or for the fact that the left-hand side depends only on KwK. Note that [w] = 1 if and only if w K. In this case it seems (empirically) that x H σ sgn x P x, w0 (q) = q l(w 0σ) σ S N. I do not have a good guess for x HwH sgn x P x, w 0 (q) in general.

27 Another equivalent formulation of our result Assume that σ σ 0, σ 0 is 213-avoiding and X σ is smooth at Y σ0. Then sgn x P x, σ (1) = sgn[w] 2 r x HwH for all w S 2N such that σ 0 w σ. Remark For σ 0 = id the condition is that X σ is smooth. Equivalently, σ is 4231 and 3412 avoiding Lakshmibai Sandhya (1990).

28 Odds and ends Suppose that X σ is smooth. Then, x HwH sgn x P x, σ (1) = sgn[w] 2 r. (1) Can one see it directly? The permutation σ (if σ id) is never (3412) avoiding, and therefore Lascoux s formula for the KL polynomials (1995, following Lascoux Schützenberger 1981) is unfortunately not applicable. However, following Deodhar (1990), it is also easy to compute P w,w if the Bott Samelson resolution of X w is small. This condition was characterized combinatorially by Billey Warrington (2001). It holds if and only if w avoids the patterns (321), ( ), ( ), ( ), ( ). In particular, for σ, this is the case if and only if σ avoids (321) and (3412). These are precisely the products of distinct simple reflections. They are all smooth, and their number is F 2N 1 where F n is the n th Fibonacci number (Fan 1998). It is a not difficult to check combinatorially that (1) holds for these σ s.

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