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Kostant's problem for fully commutative permutations

We give a complete combinatorial answer to Kostant's problem for simple highest weight modules indexed by fully commutative permutations. We also propose a reformulation of Kostant's problem in the context of fiab bicategories and classify annihilators of simple objects in the principal birepresentations of such bicategories generalising the Barbasch--Vogan theorem for Lie algebras.


Introduction and description of the results
Kostant's problem, as defined and popularized by Joseph in [Jo80], is a famous open problem in representation theory of Lie algebras. It asks for which simple modules L over a semi-simple complex finite dimensional Lie algebra the universal enveloping algebra surjects onto the space of adjointly finite linear endomorphisms of L. Although several (both positive and negative) results are known, see Subsection 3.6 for a historical overview, the general case is wide open even for simple highest weight modules in the principal block of BGG category O.
Knowing the answer to Kostant's problem for a particular simple module is important for potential applications, for example, understanding the structure of induced modules, see [KhM04,MS08], using the approach of establishing equivalences of categories for Lie algebra modules proposed in [MiSo97].
As described in Subsection 3.6, the existing results on Kostant's problem can be divided into four classes: • there are some classes of simple modules for which the answer is known to be positive; • there are some classes of simple modules for which the answer is known to be negative; • there are some results that relate the answer for one simple module to the answer for another simple module; • there are some complete classification results for Lie algebras of small rank.
To the best of our knowledge, none of the existing results provides a complete solution to Kostant's problem for some natural general class of simple modules which contains both simple modules with positive and negative answers. The main result of the present paper is the first result of this kind.
We look at a natural family of permutations (elements of the symmetric group S n ) which are called fully commutative. The latter means that any two reduced expressions for such an element can be transformed one into another just by using the commutativity relations for simple reflections. The number of fully commutative elements in S n is given by the Catalan number C n = 1 n+1 2n n . Under the Robinson-Schensted correspondence, fully commutative elements are exactly those permutations which correspond to two-row partitions. From the point of view of the Kazhdan-Lusztig combinatorics in type A, it is well known that fully commutative elements form a union of two-sided Kazhdan-Lusztig 1 cells and that, for general n, the value of Lusztig's a-function on fully commutative elements in S n can be arbitrarily large. This demonstrates that the class of fully commutative permutations is quite large. Fully commutative permutations naturally index a basis in the Temperley-Lieb algebra, which is a certain quotient of the Hecke algebra of S n .
The main result of the present paper, Theorem 5, gives a complete combinatorial classification of the fully commutative permutations such that Kostant's problem for the corresponding simple modules in the principal block of BGG category O has positive answer (we call such elements Kostant positive). The answer is both quite beautiful and rather non-trivial. We first define certain special fully commutative involutions. These are permutations of the following form: For a special involution, we call the area where this involution acts non-trivially its support. In Theorem 5, we show that a fully commutative involution w is Kostant positive if and only if it is a product of distant special involutions, in the sense that the supports of any two special factors in w are separated in the diagram of w by at least one vertical line.
Our approach to Kostant's problem for fully commutative elements is crucially based on the availability of the following three things: • the explicit description of the Serre subcategory of category O generated by the simple objects indexed by fully commutative elements and the action of projective functors on this category, due to Brundan and Stroppel, see [BS10]; The idea behind the proof is to use Kåhrström's combinatorial reformulation of Kostant's problem to reduce it to the question of solvability of certain equations in the Temperley-Lieb algebra. Then we establish, in the appropriate cases, the solvability of these equations by explicitly giving the solutions. In the remaining cases, we establish the impossibility of solving these equations by a rather technical case-by-cases analysis.
Our answer to Kostant's problem for fully commutative elements is explicit enough to compute its asymptotic behavior. More precisely, in Theorem 11, we prove the following: • asymptotically, the answer to Kostant's problem for fully commutative elements is almost surely negative; • asymptotically, the answer to Kostant's problem for fully commutative involutions is almost surely negative; • asymptotically, the answer to Kostant's problem for fully commutative elements (or involutions) of a fixed a-value is almost surely positive.
From its very definition, Kostant's problem is intimately connected with the problem of understanding the annihilators of simple modules over semi-simple Lie algebras. These annihilators are classified combining two classical theorems of Duflo and Barbasch-Vogan. In the last section of the paper, Section 7, we provide an analogue of the Barbasch-Vogan Theorem for a class of bicategories known as fiab bicategories. We also propose a reformulation of Kostant's problem in this more general context. Let g denote a semi-simple finite dimensional Lie algebra with a fixed triangular decomposition Here h is a Cartan subalgebra. We denote by W the corresponding Weyl group, by S the set of all simple reflections in W associated to the triangular decomposition (1) and by w 0 the longest element of W . Up to isomorphism, the simple objects in O 0 are the simple highest weight modules L w := L(w · 0), where 0 ∈ h * is the zero weight and w ∈ W . Here · is the usual dot-action of W on h * defined, for λ ∈ h * , via w · λ = w(λ + ρ) − ρ, where ρ is half of the sum of all positive roots.
For w ∈ W , we denote by ∆ w the Verma cover and by P w the indecomposable projective cover of L w . Let A be the opposite of the endomorphism algebra of the direct sum of the P w , taken over all w ∈ W . Then A is a finite dimensional associative algebra and the category A-mod of finite dimensional A-modules is equivalent to O 0 .
2.4. Projective functors. For w ∈ W , we denote by θ w the unique, up to isomorphism, indecomposable projective endofunctor of O 0 , in the sense of [BGG80]. It is fully determined by the property θ w P e ∼ = P w .
We denote by P the monoidal category of all projective functors acting on O 0 . Thanks to Soergel's combinatorial description, P is equivalent to the monoidal category of Soergel bimodules over the coinvariant algebra associated to W .
2.5. Graded lift. The algebra A is Koszul, in particular, it has the corresponding Koszul Z-grading, see [So90,St90]. We denote by O Z 0 the category of all finite dimensional Z-graded A-modules.
All structural modules in O 0 admit graded lifts. For indecomposable modules, these graded lifts are unique, up to isomorphism and grading shift. For indecomposable structural modules, there is a preferred graded lift, called standard lift.
Abusing notation, we will denote the ungraded objects in O 0 by the same symbols as their standard graded lifts. We use a similar convention for the graded version of projective functors, see [St90], and denote by P Z the monoidal category of all graded projective functors acting on O Z 0 . Conjecture 1. If g ∼ = sl n , then, for any x, y ∈ W , the module θ x L y is either indecomposable or zero.
We will denote by KM(x, y) ∈ {true, false} the value of the claim "the module θ x L y is either indecomposable or zero". We also denote by KM( * , y) the conjunction of the KM(x, y) over all x ∈ W .
3. Kostant's problem 3.1. Annihilators of simple modules. Let L be a simple g-module. By Dixmier-Schur's Lemma, L has central character, say χ : Z(g) → C, where Z(g) is the center of the universal enveloping algebra U (g) of g. Assume that L is such that χ coincides with the central character of the trivial g-module.
Then Duflo's Theorem [Du77] asserts that there exists w ∈ W such that the annihilator Ann U(g) (L) of L in U (g) coincides with Ann U(g) (L w ).
Further, a result of Barbasch and Vogan [BV82,BV83] says that, given x, y ∈ W , we have Ann U(g) (L x ) ⊂ Ann U(g) (L y ) if and only if x ≥ L y.

3.2.
Harish-Chandra bimodules. Recall, see [Ja83, Kapitel 6], that a g-g-bimodule is called a Harish-Chandra bimodule if it is finitely generated and, additionally, if the adjoint action of g on it is locally finite and has finite multiplicities. 3.3. Classical Kostant's problem. Kostant's problem, as advertised in [Jo80], is formulated as follows: Kostant's Problem. For which w ∈ W is the embedding an isomorphism?
We will denote by K(w) = K g (w) ∈ {true, false} the logical value of the claim "the embedding U (g)/Ann U(g) (L w ) ֒→ L(L w , L w ) is an isomorphism". Conjecture 2. Assume that we are in type A n−1 , so W ∼ = S n , the symmetric group. Let d ∈ S n be an involution. Then the following assertions are equivalent: (1) K(d) = true.
(2) For any x, y ∈ W such that x = y and θ x L d = 0 and θ y L d = 0, we have (3) For any x, y ∈ W such that x = y and θ x L d = 0 and θ y L d = 0, we have
3.6. Known results on Kostant's problem. Here is a (possibly incomplete) list of known results on Kostant's problem, in particular, a list of the special cases in which the answer to Kostant's problem is known.
• In type A, the value K(w) is constant on Kazhdan-Lusztig left cells, see [MS08, Theorem 61].
• Let p be a parabolic subalgebra of g containing h ⊕ n + . Let W p be the corresponding parabolic subgroup of W and w p 0 be the longest element in W p . Then • A complete answer to Kostant's problem for sl n , where n = 2, 3, 4, 5, is given in [KaM10, Section 4]. There one can also find a partial answer for sl 6 . Some further cases for sl 6 are dealt with in [Ka10, Section 6]. A complete answer to Kostant's problem for sl 6 is given in [KMM20, Subsection 10.1].

Fully commutative permutations and the Temperley-Lieb algebra
4.1. Fully commutative permutations. Consider the symmetric group S n as a Coxeter group in the usual way, that is, fixing the elementary transpositions s i := (i, i + 1), for i = 1, 2, . . . , n − 1, as the set of simple reflections. The classical presentation of S n with respect to these generators has the following relations: The classical result of Matsumoto [Mat64] asserts that, for w ∈ W , any reduced expression of w can be transformed into any other reduced expression of w by using only the relations in (3) and (4). An element w ∈ S n is said to be fully commutative provided that any reduced expression of w can be transformed into any other reduced expression of w by only using relation (3).
The classical Robinson-Schensted correspondence, see [Sc61] and [Sa01, Section 3.1], assigns to an element of S n a pair of standard Young tableaux of the same shape (that shape is a partition of n). An element w ∈ S n is fully commutative provided that the shape of the tableaux in RS(w) is a partition with at most two parts. Alternatively, an element w is fully commutative provided that it avoids the (3, 2, 1)-pattern, that is, the longest decreasing subsequence for w has length at most two. We refer to [Fa95,Fa96] for details.

4.2.
Temperley-Lieb algebra. Let A be a commutative ring and δ ∈ A. For n ≥ 1, the corresponding Temperley-Lieb algebra TL n (A, δ) is an A-algebra which is free as an Amodule with a basis consisting of all planar (i.e. non-crossing) pairings of 2n points in a plane. Composition is defined via concatenation of diagrams, straightening the outcome to a new diagram and, finally, multiplication with δ k , where k is the number of closed loops removed during the straightening procedure. We refer to [Mar91, Chapter 6] for details. Here is an example: one can give the following alternative description of TL n (A, δ). It is the quotient of the Hecke algebra H n for S n modulo the ideal generated by all H w , where w is not fully commutative. We note that the latter ideal is just the A-span of all such H w . By construction, TL n (Z[v, v −1 ], v +v −1 ) has a basis given by all H w , for w fully commutative. Setting e i := H si , for i = 1, 2, . . . , n − 1, we have the following presentation for Here e i corresponds to the diagram where the horizontally paired points are i and i + 1.
For a fully commutative element w ∈ W , we denote by e w the image of H w in . Taking any reduced expression w = s 1 s 2 . . . s k , we have e w = e s1 e s2 . . . e s k , see [Fa95].
The number of cups (or caps) in e w coincides with the value of Lusztig's a-function on w. Also, the cups and caps determine the corresponding Kazhdan-Lusztig right and left pre-orders, respectively. For example, x ≤ L y if and only each cap of e x is a cap of e y .
Remark 3. Let x, y, z ∈ S n be fully commutative elements such that e x e y = f (v)e z , for some f (v) ∈ A. Then the number of cups (resp. caps) in e z is greater than or equal to the maximum of the numbers of cups (resp. caps) in e x and e y .
Remark 4. Let x ∈ S n be a fully commutative element. Then e x −1 is obtained from e x by reflection in a horizontal line.

4.3.
Results of Brundan and Stroppel. Given two arbitrary fully commutative elements x, y ∈ S n , the corresponding module θ x L y in category O for sl n is either indecomposable or zero. If y belongs to a Kazhdan-Lusztig right cell containing an element of the form w p 0 w 0 , for some parabolic p, this follows by combining [BS10, Theorem 4.11] and [BS11, Theorem 1.1]. Using [MS08, Proposition 35], we can remove this assumption on the right cell of y. In particular, this implies that KM( * , w) = true for any fully commutative w ∈ S n .
Note that, by [Ge06, Theorem 5.1], the two-sided order on S n is given by the dominance order on partitions. Consequently, if x ∈ S n is not fully commutative while y ∈ S n is, then θ As a permutation, the element σ i,j has the following diagram: It is readily seen that this element is (3, 2, 1)-avoiding and hence fully commutative. The corresponding Temperley-Lieb diagram is: We will call the set {i − j, i − j + 1, . . . , i + j + 1} the support of σ i,j and the set Elements of the form σ i,j will be called special.
We will say that σ i,j and σ i ′ ,j ′ are distant provided that their extended supports intersect in at most one element. Proof. This follows from Theorem 5, the first bullet in Subsection 3.6 and the paragraph before Remark 3.
The two directions of Theorem 5 will be proved in Subsections 5.4 and 5.5.

Auxiliary lemmata.
Lemma 7. Let x and y be two fully commutative elements in S n . Then θ x L y = 0 is equivalent to the condition that each cap of e x corresponds to a cup of e y .
Lemma 8. Let x, y, z be three fully commutative elements in S n such that Then θ x θ y is isomorphic to θ f (1) z ⊕θ, where θL w = 0 for any fully commutative element w ∈ S n .
Proof. This follows from the realization of the Temperley-Lieb algebra as a quotient of the Hecke algebra and the action of the latter on O 0 via projective functors. 5.4. Positive answer. Let d be a fully commutative involution which is a product of pairwise distant special elements. We are going to prove that d has the property described in Conjecture 2(4).
Recall that, for fully commutative elements, assertions (1) and (2) in Conjecture 2 are equivalent. We also have the obvious implications (4)⇒(3)⇒(2). Hence K(d) = true for any fully commutative involution d that is a product of pairwise distant special elements.
Let x, y ∈ S n be two different fully commutative elements such that θ x L d = 0 and θ y L d = 0. To prove that d has the property described in Conjecture 2(4), it is enough to find some u, v ∈ S n such that dim Hom g (θ u P v , θ x L d ) = dim Hom g (θ u P v , θ y L d ).
By adjunction, we have Note that, for a projective module P , the dimension of Hom g (P, L d ) equals the multiplicity of P d as a direct summand of P . Therefore, we need to find u and v such that the multiplicity of θ d as a summand of θ x −1 θ u θ v is different from the multiplicity of θ d as a summand of θ y −1 θ u θ v .
Since θ d is self-adjoint and both u and v are arbitrary, we can reformulate this as follows: find u and v such that the multiplicity of θ d as a summand of θ v θ u θ x differs from the multiplicity of θ d as a summand of θ v θ u θ y .
Without loss of generality, we may assume that the number a of caps in e x is greater than or equal to the number b of caps in e y .
In most cases below, we will see that the choice v = d and u = x −1 does the job.
Case 1. Let us assume that a > b, so there is at least one cup of e x which is not a cup of e y . Set u = x −1 and v = d.
The diagram e x −1 e x is self-dual (i.e. symmetric with respect to reflection in a horizontal line), has a cups and a caps, and also a circles in the middle before straightening. By Lemma 7, each cup of this diagram corresponds to a cap of e d . Therefore, the number of circles removed when straightening the product e d e x −1 e x equals 2a. This means that the multiplicity of θ d as a summand of θ d θ x −1 θ x equals 2 2a .
If the underlying diagram of e d e x −1 e y is not e d , then, by Lemma 8, the multiplicity of θ d as a summand of θ d θ x −1 θ y equals 0 and we are done. If the underlying diagram of e d e x −1 e y is e d , we need to compute the number of closed loops removed when straightening the product e d e x −1 e y . Each closed loop contains at least one cap. The total number of original caps in e d e x −1 e y before straightening is a + b + c, where c is the number of caps in e d . As c ≥ a > b by Lemma 7, the resulting diagram has at least c caps, see Remark 3. Therefore, the number of closed loops removed during the straightening procedure is at most a + b < 2a. Consequently, the multiplicity of θ d as a summand of θ d θ x −1 θ y is 2 a+b < 2 2a . This completes Case 1.
Case 2. Let us assume that a = b and e x and e y have exactly the same cups. Set u = x −1 and v = d.
Since x and y are different by assumption, there should be at least one cap of e x which is not a cap of e y . This implies that the underlying diagrams e q of e x −1 e x and e p of e x −1 e y are different. Now we claim that the underlying diagram for e d e x −1 e x is e d , while the underlying diagram for e d e x −1 e y is different from e d . This, of course, will complete the present case. The underlying diagram of e d e x −1 e x being e d follows directly from the fact that each cap of e x is also a cap of e d , see Lemma 8. Here is an example: Let us now look at the underlying diagram for e d e x −1 e y . Recall that: • e p and e q have exactly the same cups; • not all caps in e p and e q are the same; • any cap of e p and any cap of e q is also a cap of e d ; • d = d 1 d 2 · · · d r is a product of pairwise distant special elements d i .
This implies that there exists a special factor d i of d such that the number of caps e di shares with e q is different from the number of caps e di shares with e p .
Take the leftmost such factor. Due to this assumption, to the left of this factor in e d , the diagrams e d e x −1 e x and e d e x −1 e y fully agree. We have two subcases.
Subcase 2a. The number of caps e di shares with e q is smaller than the number of caps e di shares with e p .
In this case, the additional caps of e di , when multiplied with e p , are moved to the right.
Here is a fairly generic example: This makes the resulting diagram different from e d , as claimed.
Subcase 2b. The number of caps e di shares with e q is greater than the number of caps e di shares with e p .
In this case, we have some additional points corresponding to propagating lines to the left of the nested caps in e p . The rightmost of these points either hits a cap in e d or it hits a propagating line in e d . In the former case, we obtain a cap in e d e p which is not a cap in e d . Here is a fairly generic example: This makes the resulting diagram different from e d .
In the latter case, the resulting diagram has a non-vertical propagating line. Here is a fairly generic example: This makes the resulting diagram again different from e d and completes Case 2.
Case 3. Let us assume that a = b and e x and e y have exactly the same caps. Set u = x −1 and v = d as before.
Consider first the situation when the underlying diagrams of e x −1 e x and e x −1 e y coincide, let us call this diagram e z . Note that e z has a caps. As e x and e y have the same caps but are assumed to be different, not all of their cups can coincide. In particular, the multiplicity of e z in e x −1 e y is strictly smaller than the multiplicity 2 a of e z in e x −1 e x . Consequently, if we now multiply with e d on the left, we obtain e d as the underlying diagram in both e d e x −1 e x and e d e x −1 e y , but with different multiplicities, and we are done with this situation. Take now a cap C in e d which is not nested inside any other cap in e d . Due to our assumptions on d, the immediate outside neighbors of C are propagating lines. Therefore the endpoints of C correspond to either a cup or two propagating lines in e p . In the former case, all caps nested inside C correspond to cups of e p . In the latter case, both of these propagating lines have to be vertical, for otherwise there would exist some extra caps in e p which are not caps of e d .
We proceed by induction on the number of nested caps contained inside C which do not correspond to any cups in e p , to show that e p and e q coincide, which contradicts our assumption. If this number is zero, we are done. If it is not zero, we apply the above argument to the outermost cap nested inside C. In the end, we obtain that e p equals e q , as desired. This completes Case 3.
Case 4. Let us assume that a = b and that some of the caps and some of the cups in e x and e y are different. Set u = x −1 and v = d as before. In most situations, it is possible to adapt the argument we used in Case 3.
If e d e x −1 e x = e d e x −1 e y , then we are done. So, let us assume e d e x −1 e x = e d e x −1 e y . Note that e d e x −1 e x = 2 2a e d .
As before, if the underlying diagrams of e x −1 e x and e x −1 e y are the same, the multiplicity of e d in e d e x −1 e y is strictly smaller than 2 2a . Therefore, it remains to consider the situation when the underlying diagram e q of e x −1 e x is different from the underlying diagram e p of and e x −1 e y . If e q and e p have the same caps, we can use the argument from Case 3. In particular, we may assume that not all caps in e q and e p agree. In this case we will need to construct a modification e d ′ of e d which will take the place of e v .
The same argument as in Case 3 shows that the set of propagating lines of e d is a subset of both the set of propagating lines of e p and the set of propagating lines of e q . Furthermore, the sets of caps of e p and e q are (different) subsets of the set of caps of e d .
Let us consider some full collection F of nested caps in e d . Let α and β be the numbers of caps of e p , respectively, e q contained in this collection.
Note that e q has a caps while e p has at least a caps. Therefore, since not all caps in e p and e q agree, we can assume that α > β for the chosen F . The argument from Case 3 implies that each cap in F either hits a cup or two vertical lines in e p . This means that the parts of e p and e q corresponding to F are as in the following example: Consider the element d ′ such that e d ′ is the same as e d except F is adjusted as follows: Then e d ′ e p is a multiple of e d while e d ′ e q is not. This completes the proof for the positive answer.
5.5. Negative answer. Let d be a fully commutative involution which is not a product of pairwise distant special elements. We are going to prove that d does not have the property described in Conjecture 2(2). More explicitly, we will find two different fully commutative elements x and y such that θ x L d is isomorphic to θ y L d .
Recall that, for fully commutative elements, assertions (1) and (2) in Conjecture 2 are equivalent. We also have the obvious implications ¬(2)⇒ ¬(3)⇒ ¬(4) of the other assertions. The above therefore implies K(d) = false and, moreover, the rest of Conjecture 2 for the involution d.
If d is not a product of pairwise distant special elements, the diagram e d has two adjacent non-nested caps. Let us fix a pair A and B of such adjacent non-nested caps (with A on the left). We may assume that they are not nested in some other cap which itself has an adjacent non-nested cap. Let A ′ and B ′ be the corresponding cups.
Define the element e x by changing e d as follows: • remove B and B ′ ; • if applicable, remove all caps in which A and B are nested; • remove all cups corresponding to the latter caps; • replace all the removed cups and caps by propagating lines.
The latter process is unique due to the non-intersection condition.
Define the element e y by changing e d as follows: • remove A and B ′ ; • if applicable, remove all caps in which A and B are nested; • remove all cups corresponding to the latter caps; • replace all the removed cups and caps by propagating lines.
Again, the latter process is unique due to the non-intersection condition. Here is an example, with A and B colored: Let c denote the number of caps in e d and a the number of caps in e x . Then a ≤ c − 1, by construction, and also a equals the number of caps in e y . We are going to prove that θ x L d and θ y L d are isomorphic as graded modules. For this we need some combinatorial preparation for estimates of graded shifts. The following statements can probably be deduced from the results of [BS10], but it is easier to prove them directly.
Lemma 9. Let u and w be two fully commutative permutations such that θ u L w = 0. Let k be the minimum of the numbers of caps in e u and in e w . Then θ u L w is a graded self-dual module and, for |i| > k, the graded component (θ u L w ) i is zero.
Proof. The module θ u L w is self-dual as it is the image of a self-dual module L w under a projective functor. In order to prove the rest of the lemma, it is enough to argue that, for i < −k, the graded component (θ u P w ) i is zero. Since the algebra of O 0 is positively graded, all standard graded lifts of projectives live in non-negative degrees. When computing the product e u e w , the number of closed loops removed in the straightening procedure is at most k. This gives the scalar (v + v −1 ) m , where m ≤ k. Therefore the maximal graded shift of a projective in θ u P w is bounded by k.
The claim follows.
Corollary 10. Both modules θ x L d and θ y L d have simple tops, which live in degree −a.
Proof. Both modules θ x L d and θ y L d have simple tops by [BS10,Theorem 4.11]. We prove the second claim for θ x L d . For θ y L d the arguments are similar. Lemma 9 implies that the simple top of θ x L d lives in some degree i ≥ −a.
Let us now look more closely at the proof of Lemma 9. Note that, when computing the product e x e d , we need to remove exactly a circles. This means that the degree −a component of θ x P d is non-zero.
At the same time, any simple subquotient L w in the radical of P d lives in a strictly positive degree. Therefore, by Lemma 9, the module θ x L w lives in degrees that are strictly bigger than −a. Thus the degree −a component of θ x L d is indeed non-zero. Due to the positivity of the grading, this component is the top of θ x L d .
We want to prove that two graded modules θ x L d and θ y L d which have simple tops that live in the same degree are isomorphic. Due to the positivity of the grading, it is enough to show that there is a non-zero degree zero morphism from θ x L d to θ y L d . By adjunction, this is equivalent to the existence of a degree zero morphism from L d to By construction, e x and e y have the same cups. Therefore, during the straightening of the product e x −1 e y , there are a loops to remove. Let e p be the underlying diagram of e x −1 e y and note that it has exactly the same caps as e y . By an analogue of Corollary 10 for e p (which works verbatim since e p and e y have the same caps), the module θ p L d is a self-dual indecomposable module with simple top in degree −a. Since θ p appears exactly once with shift a in the decomposition of θ x −1 θ y , it follows that L d appears, as a graded module, in the socle of θ x −1 θ y L d . This completes the proof of the negative answer and the proof of Theorem 5.
5.6. Sanity check: comparison to previously known results. As already mentioned in Subsection 3.6, it is known that all elements of the form w p 0 w 0 , where p is a parabolic subalgebra of sl n , are Kostant positive.
If we take p to be a maximal parabolic subalgebra, that is, one for which the semisimple part of the Levi factor equals sl i ⊕ sl n−i , for i = 0, 1, . . . , ⌊ n 2 ⌋, the element w p 0 w 0 turns out to be fully commutative. The Temperley-Lieb diagram of the element w p 0 w 0 is as follows, where i is the number of caps: If i = n 2 , this element is an involution, in fact, it is σ a,b , where a = n 2 and b = n 2 − 1. If i < n 2 , the above element belongs to the left Kazhdan-Lusztig cell of σ a,b , where a = i and b = i − 1. Therefore, for such elements, our Theorem 5 agrees with the previous results.
Moreover, all known results in small ranks mentioned in Subsection 3.6 indeed agree with Theorem 5.
We also note that Theorem 5, combined with [Ka10, Theorem 1.1], gives a lot of new full answers to Kostant's problem even for not necessarily fully commutative permutations. 5.7. Problems to extend outside fully commutative elements. The fact that the answers to Kostant's problem for the elements s 1 s 2 s 1 and s 2 s 3 s 2 of S 5 are different, see [KaM10, Section 4], suggests that it will not be straightforward to extend Theorem 5 outside the set of fully commutative elements.
6. Asymptotic results 6.1. Various sequences. For n ∈ Z ≥1 , we denote • by ki n the number of fully commutative Kostant involutions in S n ; • by k n the number of fully commutative w ∈ S n for which K(w) = true; • by mi n the number of fully commutative involutions in S n ; • by m n the number of fully commutative elements in S n .
For a ∈ Z ≥1 and x ∈ {ki n , k n , mi n , m n }, we denote by x a the number of elements in the family x with exactly a caps.
The remainder of this section is devoted to the proof of this theorem. In Subsections 6.3, 6.4 and 6.5, we first establish some explicit formulae for the enumeration of the main protagonists defined in the previous subsection. We are sure that some of the combinatorial arguments and results presented in this section are not new and can be found in or derived from the existing literature. However, we feel that is would be more difficult to find appropriate references than to prove these results. When working on the proofs, the Online Encyclopedia of Integer Sequences was really helpful.
6.3. Fully commutative Kostant involutions. Recall the family of Fibonacci polynomials F n (x), where n ≥ 0, given by the following recursion: Here are some initial members of this family: The evaluation F n (1) is exactly the n-th Fibonacci number.
Proof. This is easy to check for n = 1, 2. Therefore, it suffices to show that the numbers ki a n satisfy the same recursion as the coefficients of the Fibonacci polynomials.
Let d be a Kostant involution and look at the strand in e d which starts at the top point 1. If this strand is vertical, removing it yields a Kostant involution for n − 1 with the same number of caps as e d . If this strand is a cup, removing it together with the corresponding cap produces a Kostant involution for n − 2 with one fewer cap than e d . This defines a bijection between the set of all Kostant involutions for n with a caps and the union of the set of all Kostant involutions for n − 1 with a caps and the set of all Kostant involutions for n − 2 with a − 1 caps. This implies ki a n = ki a n−1 + ki a−1 n−2 , which establishes the necessary recursion.
Corollary 13. The number ki n is the n-th Fibonacci number.
Proof. Evaluate the equality in Proposition 12 at 1.
Corollary 14. We have ki a n = n−a a .
Proof. Again, this is easy to verify for small values of n, so we only need to check that the binomial coefficients on the right hand side satisfy the same recursion as ki a n . By the usual Pascal triangle formula, we have: This implies the claim.
Proof. In type A, two-sided Kazhdan-Lusztig cells are in bijection with partitions of n, see [KL79]. The two-sided cell which corresponds to our a is indexed by the partition (n − a, a). Since every Kazhdan-Lusztig left cell contains a unique involution, the number k a n is the product of ki a n with the size of this left cell. The latter equals the number of standard Young tableaux of shape (n − a, a). Now the claim of our corollary follows from Corollary 14 and the Hook Formula. Proof. This claim is easy to check for small values of n, so we need to show that both sides satisfy the same recursion.
Let d be a fully commutative involution. Consider the strand of e d connected to the upper point 1. If it is vertical, removing it results in a fully commutative involution for n − 1. If the strand is a cup X connecting 1 to some point 2i, there is a crossingless pairing of 2i − 2 points inside this cup. Removing X and all cups contained inside it together with the corresponding caps, we obtain a fully commutative involution for n − 2i. This implies that we have the following recursion for the left hand side of our formula: We claim the the middle binomial coefficients n ⌊ n 2 ⌋ satisfy the same recursion. Indeed, consider the Pascal triangle with the point (x, y) corresponding to x y , for all appropriate x and y. Then n i is exactly the number of shortest paths between (0, 0) and (n, i) in this triangle.
Assume first that n = 2k. Then we need to prove that Equivalently, we need to show that For i = 1, 2, . . . , k, let P i be the set of all shortest paths between (0, 0) and (2k − 1, k) such that the path goes through the point (2(k − i), k − i) and i is minimal with this property. The set of all paths between (0, 0) and (2k − 1, k) is the disjoint union of the P i . There are 2k−2i k−i shortest paths between (2(k − i), k − i) and (0, 0). The classical interpretation of Catalan numbers as paths in a square that do not cross the diagonal implies that there are exactly C i−1 shortest paths between (2k − 1, k) and (2(k − i), k − i) satisfying the condition for the minimality of i. This establishes the necessary recursion formula for the right hand side.
Here both the numerator and the denominator are polynomials in n of degree a and with leading coefficient 1. The claim of Theorem 11(c) follows. 6.7. Proof of Theorem 11(a). By Corollary 13, the number ki n is the n-th Fibonacci number. It is given by the formula Since the absolute value of 1− √ 5 2 is less than 1, it follows that F n grows as ( 1+ √  5 2 ) n . At the same time, if n = 2k, the middle binomial coefficient 2k k is not smaller than , while, if n = 2k + 1, the coefficient 2k+1 k is not smaller than 2k k . This implies that n ⌊ n 2 ⌋ grows faster than 2 n n . Since 1+ √ 5 2 < 2, the claim of Theorem 11(a) follows.
6.8. Proof of Theorem 11(b). Using Corollary 15, we need to show that We rewrite the expression in (8) as (n + 1) n−a a n−a a n a n−a+1 a 2 ) n . The coefficient n a is bounded by n ⌊ n 2 ⌋ and hence grows as 2 n up to some factor of at most polynomial growth. At the same time, 2n n grows as 4 n up to some factor of at most polynomial growth. As the number of summands is linear in n, it follows that the whole expression (10) tends to 0, when n → ∞. This proves Theorem 11(b). 6.9. Conjectures. Taking Theorem 11 into account, we conjecture the following for general elements of S n : • Among all involutions in S n , the proportion of those for which the answer to Kostant's problem is positive is asymptotically 0.
• The proportion of elements in S n for which the answer to Kostant's problem is positive is asymptotically 0.
• Fix a partition λ of some m and consider, for n > m, the partition λ (n) of n obtained from λ by increasing the first part by n − m. Then, among the elements in S n belonging to the two-sided cell indexed by the partition λ (n) , the proportion of those for which the answer to Kostant's problem is positive is asymptotically 1. Denote byL the set of all F ∈ S(C ) such that F ≤ L L. The collection of Serre subcategories of the P i (j), for j ∈ C , generated by all simple objects L F , where F ∈L, is invariant under the action of C . We denote the corresponding (abelian) birepresentation of C by M L and its (finitary) restriction to projective objects by M L .
On the other hand, let D L be the Duflo 1-morphism in L and consider the finitary birepresentation K L given by the action of C on the additive closure of CL DL . The Yoneda morphism from P i to K L sending ½ i to L DL factors through M L by construction. We will say that D L is Kostant positive provided that the induced morphism of birepresentations from M L to K L is an equivalence.
In the case of the bicategory P , this is indeed equivalent to Kostant's original problem for category O by [KMM20, Corollary 7.6].
Note that the kernel of the above Yoneda morphism from P i to K L is exactly the annihilator of L DL in C , i.e. the left biideal of 2-morphisms α in C such that P i (α) LD L = 0. This connects our reformulation of Kostant's problem in this more general setup to the problem of studying annihilators of simple objects in principal birepresentations. This naturally leads to an analogue of the classical Barbasch-Vogan theorem for fiab bicategories, presented in the next subsection.
7.3. Barbasch-Vogan theorem for C. Note that the abelianization P i of P i is an abelian birepresentation of C. For j ∈ C and an indecomposable 1-morphism F ∈ C (i, j), denote by J F the annihilator of L F in C . Then J F is a left biideal of C .
Theorem 17. For two indecomposable 1-morphism F and G in C , the following conditions are equivalent: In this proof, we will use the word "module" instead of "birepresentation". The direct sum of all principal C-modules has the obvious structure of a C-C-bimodule, the regular C -C-bimodule.
Given an object M of this regular C -C-bimodule and any 1-morphism F ∈ C which acts on M on the right, we have Let F and G be two 1-morphisms in C. Then F ≤ R G if and only if there exists H in C such that G is a summand of FH. By adjunction, this is equivalent to 0 = Hom C (FH, L G ) ∼ = Hom C (F, L G H ⋆ ).
In other words, F ≤ R G is equivalent to the existence of H such that L F is a subquotient of L G H ⋆ . The previous paragraph now yields that F ≤ R G implies J G ⊂ J F .
For the converse, note that if F R G, then either F > R G, or F and G are not comparable in the right order.
If G < R F, then all elements in the two-sided cell of F annihilate L G by [MM11, Lemma 12]. By the same lemma, there are elements in the two-sided cell of F that do not annihilate L F . Therefore J F J G .
If F and G are not comparable in the right order, then the two-sided cells of F and G annihilate L G and L F , respectively, again by [MM11,Lemma 12]. On the other hand, the two sided cell of F does not annihilate L F , and similarly that of G does not annihilate L G . Thus the annihilators are not comparable.
This completes the proof.
We can also give slightly more detailed information.
Proposition 18. In the setup of Theorem 17, if F < R G and i is the codomain for both F and G, then there is H ∈ C (i, i) such that dim Hom C /J G (H, ½ i ) = dim Hom C /J F (H, ½ i ).

Proof.
We take H to be the Duflo 1-morphism in the right cell of G. Then The appearance of the left order in Corollary 19, compared to the right order in Theorem 17, is due to the right nature of the action of the bicategory of projective functors on category O.
Proof. Consider the bicategory P of projective functors acting on O 0 . The latter is, naturally, a subbirepresentation of the abelianized principal birepresentation P and is equivalent to a certain category of Harish-Chandra bimodules for g by [BGG80,Theorem 5.9]. This equivalence matches the indecomposable projective object P θe corresponding to the identity with the quotient of U (g) modulo the trivial central character.
Sending P θe to the dominant Verma module in O 0 using the Yoneda Lemma, defines a morphism of birepresentations from P to O 0 which sends simple objects to simple objects. By Theorem 17, y ≤ L x implies Ann P (L x ) ⊂ Ann P (L y ).

In particular,
Hom Ann P (Lx) w∈W θ w , θ e ⊆ Hom Ann P (Ly) w∈W θ w , θ e ⊆ Hom P w∈W θ w , θ e , where the latter corresponds to the quotient of U (g) modulo the ideal generated by the trivial central character under the equivalence from [BGG80, Theorem 5.9]. Hence Ann U(g) (L x ) ⊆ Ann U(g) (L y ).
Since P has only one object, the fact that y < L x implies Ann U(g) (L x ) Ann U(g) (L y ) follows directly from Proposition 18. As in the proof of Theorem 17, if x and y are not comparable in the left order, their annihilators are incomparable, which completes the proof.