Kazhdan-Lusztig polynomials for \tilde{B}_2
aa r X i v : . [ m a t h . R T ] F e b Kazhdan-Lusztig polynomials for e B Karina Batistelli ∗ , Aram Bingham † , and David Plaza ‡ Abstract
In their seminal paper [KL79] Kazhdan and Lusztig define, for an arbitrary Coxeter system ( W, S ) ,a family of polynomials indexed by pairs of elements of W . Despite their relevance and elementarydefinition, the explicit computation of these polynomials is still one of the hardest open problems inalgebraic combinatorics. In this paper we explicitly compute Kazhdan-Lusztig polynomials for a Coxetersystem of type e B . Kazhdan-Lusztig polynomials lie at the intersection of representation theory, geometry and algebraic com-binatorics. Their relevance derives from the fact that they elegantly express the answers to many difficultproblems in the aforementioned areas. Despite their straightforward definition (through a recursive algo-rithm involving only elementary operations) and enormous efforts made by several authors over the pastforty years, in general the explicit computation of Kazhdan-Lusztig polynomials remains elusive.In this paper we compute Kazhdan-Lusztig polynomials in type e B ; for relevant background and relatedtheory we refer the reader to [EMTW20]. In order to present our results we introduce some notation. Themain protagonist in this paper is the affine Weyl group of type e B , which we denote by W . It is a Coxetergroup generated by involutions S = { s , s , s } with relations ( s s ) = ( s s ) = ( s s ) = 1 . As usual, wedenote by ℓ ( · ) and ≤ the length and Bruhat order on W , respectively. It is convenient to recall the realizationof W as the group of isometric transformations of the plane generated by reflections in the lines that supportthe sides of an isosceles right triangle as is illustrated in Figure 1. In this figure the triangle marked witha dot represents the identity of W . The colors on the edges of the triangles represent the following simplereflections: red is s , blue is s and green is s . Given an element w ∈ W and an expression s i s i . . . s i k (notnecessarily reduced) of w we define a sequence of triangles ( △ , △ , . . . △ k ) as follows: △ is the identitytriangle. Then △ j +1 is obtained from △ j by reflecting it through the side colored with s i j . We identify w with △ k . Of course, this identification does not depend on the choice of an expression for w . Henceforth,we do not distinguish between elements of W and triangles.We split W − { } into three regions: The big region (the region colored using the lightest gray), the thickregion (the region colored by the darkest gray) and the thin region (the region colored with the intermediategray). The reader familiar with cells in Kazhdan-Lusztig theory might have noticed the small discrepancybetween Figure 1 and [Lus85, Figure 2]. The big/thick/thin regions coincide very nearly with two-sided cells,though we will prefer our description in order to keep this paper self-contained.Let H be the Hecke algebra of W with standard basis { H w } w ∈ W and Kazhdan-Lusztig basis (canonicalbasis) { H w } w ∈ W . Kazhdan-Lusztig polynomials { h x,w ( v ) | x, w ∈ W } are defined by the equality H w = X x ∈ W h x,w ( v ) H x . There is a group automorphism ϕ : W → W which interchanges s and s and fixes s . This extendsto a Hecke algebra automorphism which we also denote by ϕ . We have H ϕ ( w ) = ϕ ( H w ) for all w ∈ W . Tocondense notation we often write w ′ := ϕ ( w ) , for w ∈ W .In this paper we prove an explicit formula for H w for all w ∈ W located in either the big region or thethick region, and we conjecture explicit formulas for the thin region. The remainder of this introduction isdevoted to explaining these formulas. ∗ Universidad de Chile † Tulane University of Louisiana ‡ Universidad de Talca Figure 1: Geometric realization of W .We begin by considering elements in the big region. Let a = s s s , b = s s s , c = s s s s , and d = s s s s . Let N be the set of non-negative integers. For ( m, n ) ∈ N we define θ ( m, n ) = ( ( ab ) k a s d n , if m = 2 k ; ( ab ) k +1 s c n , if m = 2 k + 1 . (1.1)We define t m = s for m even and t m = s for m odd. Then (modulo ϕ ) all the elements in the big regionare of the form xθ ( m, n ) y , where x ∈ { , s , s s , s s s } and y ∈ { , t m , t m s , t m s t ′ m } .For ( m, n ) ∈ N we define Supp( m, n ) = (cid:26) { ( m − i, n − j ) ∈ N | i, j ∈ N } , if m is odd; { ( m − i, n − j ) ∈ N | i, j ∈ N } − { (0 , b ) | b n mod 2 } , if m is even.For w ∈ W we define N w = X x ≤ w v l ( w ) − l ( x ) H x . Theorem 1.1.
Let ( m, n ) ∈ N . Then, we have H θ ( m,n ) = X ( a,b ) ∈ Supp( m,n ) v ( m − a )+2( n − b ) N θ ( a,b ) . Furthermore, if x ∈ { , s , s s , s s s } and y ∈ { , t m , t m s , t m s t ′ m } then H xθ ( m,n ) y = X H θ ( m,n ) Y , where X = , if x = 1; H s , if x = s ; H s H s − , if x = s s ; H s H s H s − H s − H s , if x = s s s ; Y = , if y = 1; H t m if y = t m ; H t m H s − if y = t m s ; H t m H s H t ′ m − H t m − H t ′ m if y = t m s t ′ m . We now move to the thick region. We denote by N (resp. S , E and W ) the sub-region of the thick regionformed by triangles located to the north (resp. south, east and west) of the identity triangle. Consider theinfinite sequence { a n } ∞ n =1 = ( s , s , s , s , s , s , s , s , s , s , s , s , . . . ) . (1.2)We define x n = a · · · a n and x n = s s s x n − . Then, N = { x n | n ≥ } ∪ { x n | n ≥ } . In what follows werefer to N as the north wall. We notice that S = ϕ ( N ) . We define e n = s x ′ n . Then, E = { e n | n ≥ } ∪ { e ′ k | k ≥ } . Finally, we define w n = s e n . Then, we have W = { w n | n ≥ } ∪ { w ′ k | k ≥ } ∪ { s , s s , s s } . N . As we already pointed out S = ϕ ( N ) . Therefore, the formulas for elements located in S are obtainedby applying ϕ to the formulas in the theorem. For the sake of brevity, in this introduction we omit theformulas for the elements located in E and W . These formulas are presented in §5.2. Theorem 1.2.
For all k ≥ we have H x f ( k ) = N x f ( k ) + v N x f ( k − + k − X j =2 v j − ( N e f ( k − j ) + N u f ( k − j ) ) + v k − N s s , (1.3) where f ( k ) := 3 k + 1 and u n = s x n . Furthermore, using the convention of Remark 3.8 we have H x k +1 H s = H x k +2 , (1.4) H x k +2 H s = ( H x k +3 + H x k +1 + H θ ( k − , + H s θ ′ ( k − , + H θ ( k − , , if k is even; H x k +3 + H x k +1 + H s s s θ ( k − , , if k is odd. H x k +2 H s = ( H x k +3 + H x k +1 + H s s s θ ( k − , , if k is even; H x k +3 + H x k +1 + H θ ( k − , + H s θ ′ ( k − , + H θ ( k − , , if k is odd. (1.5)Conjectural formulas for H w for w located in the thin region are presented in §6. It is likely that theycould be proved by the methods used to obtain formulas in the other regions, though the accounting becomesmore difficult due to the number and type of terms appearing in the explicit formulas. We leave this taskfor a future investigation.The alert reader might have noticed that the results presented in this introduction provide formulas forKazhdan-Lusztig basis elements rather than Kazhdan-Lusztig polynomials. By the nature of our formulas,in order to compute Kazhdan-Lusztig polynomials we need to understand the elements N w , or, equivalently,to understand the sets ⋖ w := { x ≤ w | x ∈ W } . This is exactly the content of §2. We will finish thisintroduction with an example that illustrates how our formulas can be used in order to efficiently computeKazhdan-Lusztig polynomials.Suppose we want to compute h x n ,x m ( v ) for integers m > n ≥ , m odd and n even. We notice thattaking the coefficient of H x n on both sides of (1.5) and using (3.4) we obtain v − h x n ,x m − ( v ) + h x n − ,x m − ( v ) = h x n ,x m ( v ) + h x n ,x m − ( v ) + h x n ,s s s θ ( m − , ( v ) . On the other hand, (1.4) and (3.4) imply h x n ,x m − ( v ) = vh x n ,x m − ( v ) + h s s s θ ( n − , ,x m − ( v ) ,h x n − ,x m − ( v ) = v − h x n − ,x m − ( v ) + h x n − ,x m − ( v ) . For an integer l we define F l ( v ) = l − X i =0 v i .We can use (1.3) and the description of lower intervals in §2 to obtain h x n ,x m − ( v ) = v m − n ( F m − n ( v ) + F m − n − ( v )) ,h s s s θ ( n − , ,x m − ( v ) = v m − n +1 ( F m − n − ( v ) + F m − n − ( v )) ,h x n − ,x m − ( v ) = v m − n +1 ( F m − n ( v ) + F m − n − ( v )) ,h x n − ,x m − ( v ) = v m − n ( F m − n +1 ( v ) + F m − n − ( v )) . Similarly, using Theorem 1.1 (or Lemma 4.13) we obtain h x n ,s s s θ ( m − , ( v ) = v m − n ( F m − n ( v ) + F m − n − ( v )) . Putting all these together we obtain h x n ,x m ( v ) = v m − n ( F m − n +1 ( v ) + 2 F m − n − ( v ) + F m − n − ( v )) . (1.6)3losed formulas for h x,w ( v ) , such as (1.6), can be obtained using the results in this paper for all x ∈ W andfor all w ∈ W located either in the big region or the thick region. We have chosen this specific examplein order to point out a certain discrepancy between an identity appearing in [Lus97, (a)] and our results.Concretely, in that paper it is claimed that µ ( x n , x m ) = 1 , (1.7)for all m > n > , m odd, n even, where µ ( x, w ) denotes the coefficient of v in the Kazhdan-Lusztigpolynomial h x,w ( v ) . However, it is clear from (1.6) that µ ( x n , x m ) = 0 if m − n > . Equation (1.7)allows the author to conclude that the W -graph of an affine Weyl group of type e B is not locally finite.This conclusion was later shown to be true by Wang [Wan11] through a different set of examples which areconsistent with our results.This paper is organized as follows. In Section 2 we describe geometrically the lower intervals in theBruhat order of the elements of the big region and the thick region. We specify the region correspondingto each lower interval inductively and obtain closed formulas for their sizes. In Section 3 we introduce theHecke algebra of W and prove several technical lemmas containing identities which will be necessary for thecomputations that follow. Our main results lie in Sections 4 and 5, which provide explicit formulas for theKazhdan-Lusztig basis elements of elements in the big region and the thick region respectively. Finally, inSection 6 we conjecture the explicit formulas of the Kazhdan-Lusztig basis elements corresponding to theelements of the thin region. Acknowledgements
The first author is supported by Fondecyt project 3190144. The third author was partially supported byFondecyt project 1200341. The authors would like to thank Nicolas Libedinsky for useful discussions andall the participants of the "Soergel bimodules learning seminar", which took place at the Universidad deChile during 2020, and from which this project and collaboration was born. We also wish to acknowledgethe contributors and developers of SageMath [Sag20] of which we have made indispensable use.
Throughout this paper W denotes the affine Weyl group of type e B with generators S = { s , s , s } . In thissection we provide a geometric description for lower Bruhat intervals and obtain formulas for their size. Given w ∈ W we define ≤ w := { x ∈ W | x ≤ w } and | w | := |≤ w | . We also define D R ( w ) := { s ∈ S | ws < w } and D L ( w ) = D R ( w − ) . We recall from (1.1) the definition of the elements θ ( m, n ) . We notice that D L ( θ ( m, n )) = { s , s } and D R ( θ ( m, n )) = { s , t m } . In particular, D R ( θ ( m, n )) only depends on the parity of m . Left and right descentsets characterize θ -elements in the following sense: If x ∈ W satisfies D L ( x ) = { s , s } and D R ( x ) = { s , t } for some t ∈ { s , s } , then there exists ( m, n ) ∈ N such that x = θ ( m, n ) .All these θ -elements belong to the same connected component of the big region. We denote this connectedcomponent by C . All of the elements in C are of the form θ ( m, n ) x for some x ∈ { , t m , t m s , t m s t ′ m } . Lemma 2.1.
The set ≤ θ ( m, is the square S ( m, defined as the smallest square satisfying the followingproperties: • S ( m, contains θ ( m, . • The center of S ( m, , say O , is the upper vertex of the identity triangle. • The two diagonals of S ( m, are the horizontal and vertical lines passing through O . We have slightly modified the original statement in order to match the conventions in this paper. We remark that throughout this paper we follow Soergel’s normalization in [Soe97]. Figure 2: Lower intervals: From lightest to darkest gray θ (0 , – θ (4 , . Darkestyellow θ (4 , and lightest yellow θ (4 , . Proof.
We proceed by induction on m . For m = 0 the claim is clear. We now assume the lemma holds forsome m ≥ . Note that θ ( m +1 ,
0) = θ ( m, t m s t m . Let w ≤ θ ( m +1 , . Then w = xy for some x ≤ θ ( m, and y ≤ t m s t m . By our inductive hypothesis we know that x ∈ S ( m, . It follows that w ∈ S ( m + 1 , .Therefore, ≤ θ ( m + 1 , ⊆ S ( m + 1 , . Conversely, let w ∈ S ( m + 1 , . If w ∈ S ( m, we are done, so wecan assume that w S ( m, . It is easy to see that w can be obtained from some x ∈ S ( m, by reflecting italong one of the following sequences: t m , t m s or t m s t m . Once again, our inductive hypothesis guaranteesthat w ≤ θ ( m + 1 , . Thus S ( m + 1 , ⊆ ( ≤ θ ( m + 1 , .Any square obtained from S (0 , by a translation will be called a B -square. Let m, n ∈ N . Werecursively define sets S ( m, n ) as follows: If n = 0 then S ( m, is defined as in Lemma 2.1. Assume that S ( m, n ) has been defined. Then S ( m, n + 1) is defined as the set obtained from S ( m, n ) by surroundingits boundary with B -squares. This is illustrated in Figure 2. Starting from S (4 , (the black square), weobtain S (4 , by adding B -squares (in dark yellow). Then, we obtain S (4 , by adding B -squares(in light yellow). Definition 2.2.
Let x ∈ W . We define ∂ ( x ) to be the set formed by all the elements y ≤ x such that thetriangle associated to y has a side belonging to the boundary of ≤ x . We partition ∂ ( x ) into three (possiblyempty) sets according to the color of the relevant side. More precisely, we define ∂ s k ( x ) to be the set of all y ∈ ∂ ( x ) such that the side of y belonging to the boundary of ≤ x is colored by s k . Definition 2.3.
For any
X, Y ⊆ W we define XY := { xy | x ∈ X, y ∈ Y } . In case X = { x } , we may alsowrite x ( Y ) to mean the same. Lemma 2.4.
For all m, n ∈ N we have ≤ θ ( m, n ) = S ( m, n ) .Proof. We proceed by induction on n . The case n = 0 is covered by Lemma 2.1. Assume the lemma holdsfor some fixed n . We have θ ( m, n + 1) = θ ( m, n ) t m s t ′ m s and this implies ≤ θ ( m, n + 1) = ( ≤ θ ( m, n ))( ≤ t m s t ′ m s ) . (2.1)From this it is easy to see, using our inductive hypothesis, that ( ≤ θ ( m, n + 1)) ⊆ S ( m, n + 1) .Conversely, we have S ( m, n + 1) = S ( m, n ) ∪ [ x ∈ X ∂ ( θ ( m, n )) x, where X = { t m , t m s , t m s t ′ m , t m s t ′ m s } . It follows from our inductive hypothesis and (2.1) that S ( m, n +1) ⊆ ( ≤ θ ( m, n + 1)) . We conclude that S ( m, n + 1) = ( ≤ θ ( m, n + 1)) as desired.5 orollary 2.5. For all m, n ∈ N we have | θ ( m, n ) | = 8( m + 2 m + 4 mn + 2 n + 2 n + 1) . In particular, wehave | θ ( m, | = 8( m + 1) .Proof. The equality follows easily from Lemma 2.1 and Lemma 2.4 by a counting argument. Indeed, ≤ θ ( m, is made of ( m + 1) (non-intersecting) B -squares, which leads to |≤ θ ( m, | = 8( m + 1) . Furthermore,the number of B -squares that one needs to add to ≤ θ ( m, in order to obtain ≤ θ ( m, n ) is m + 1) n +2 n ( n − . • (a) From lightest to darkest grey: Lower intervals of θ (2 , , θ (2 , s , θ (2 , s s and θ (2 , s s s . • (b) From lightest to darkest grey: Lower intervals of s θ (2 , , s θ (2 , s , s θ (2 , s s and s θ (2 , s s s . • (c) From lightest to darkest grey: Lower intervalsof s s θ (2 , , s s θ (2 , s , s s θ (2 , s s and s s θ (2 , s s s . • (d) From lightest to darkest grey: Lower intervals of s s s θ (2 , , s s s θ (2 , s , s s s θ (2 , s s and s s s θ (2 , s s s . Figure 3: Lower intervals in the big region.We now continue with the description of the sets ≤ θ ( m, n ) t m , ≤ θ ( m, n ) t m s and ≤ θ ( m, n ) t m s t ′ m .In order to obtain ≤ θ ( m, n ) t m , we take as a starting point ≤ θ ( m, n ) = S ( m, n ) and we add to it thetriangles that have an adjoining side matching the color of t m . Similarly, we can obtain ≤ θ ( m, n ) t m s from ≤ θ ( m, n ) t m and ≤ θ ( m, n ) t m s t ′ m from ≤ θ ( m, n ) t m s . In formulas we have ≤ θ ( m, n ) t m = ≤ θ ( m, n ) ⊔ ∂ t m ( θ ( m, n )) t m . (2.2) ≤ θ ( m, n ) t m s = ≤ θ ( m, n ) t m ⊔ ∂ s ( θ ( m, n ) t m ) s ; (2.3) ≤ θ ( m, n ) t m s t ′ m = ≤ θ ( m, n ) t m s ⊔ ∂ t ′ m ( θ ( m, n ) t m s ) t ′ m . (2.4)This construction is illustrated in Figure 3a for θ (2 , . With this description in hand we are in positionto obtain the sizes of these sets. 6 emma 2.6. Let m, n ∈ N . Then we have | θ ( m, n ) t m | = 8( m + 3 m + 4 mn + 2 n + 4 n + 2) , (2.5) | θ ( m, n ) t m s | = 8( m + 4 m + 4 mn + 2 n + 5 n + 3) , (2.6) | θ ( m, n ) t m s t ′ m | = 8( m + 5 m + 4 mn + 2 n + 6 n + 4) . (2.7) Proof.
It is a straightforward counting exercise to show that | ∂ t m ( θ ( m, n )) | = 8( m + 2 n + 1) . Thus (2.5)follows from Corollary 2.5 and (2.2). Similarly, we have that | ∂ s ( θ ( m, n ) t m ) | = 8(2 m + 3 n + 2) . This gives us(2.6) by combining (2.3) and the already proved identity (2.5). Finally, we obtain (2.7) by combining (2.4)and the already proved identity (2.6), together with the equality (cid:12)(cid:12) ∂ t ′ m ( θ ( m, n ) t m s ) (cid:12)(cid:12) = 8(3 m + 4 n + 1) .We now move on to the description of the remaining components of the big region. We begin with theregion s C . Let w = s x for some x ∈ C . Then ≤ w = ≤ x ∪ s ( ≤ x ) . Geometrically, this implies that ≤ w can be obtained as the union of ≤ x with the image of ≤ x under the reflection through the green line thatsupports a side of the identity triangle. Since we have already obtained a geometric description of all thesets ≤ x for x ∈ C , we now have a geometric description of all the sets ≤ w for w ∈ s C . Examples are givenin Figure 3b, obtained using those from Figure 3a.We now describe lower intervals for elements in s s C . Let w ∈ s s C . Then w = s x for some x ∈ s C .Arguing as in the previous paragraph, we get that ≤ w is the union of ≤ x with the image of ≤ x under thereflection through the blue line that supports a side of the identity triangle. This is illustrated in Figure 3cusing the examples from Figure 3b.Finally, we describe the lower intervals for elements in s s s C . Let w = s x for some x ∈ s s C . Thistime ≤ w corresponds to the union of ≤ x with the image of ≤ x under the reflection through the red linethat supports a side of the identity triangle. We have illustrated this case in Figure 3d starting from thecorresponding pictures in Figure 3c.Having described the lower intervals geometrically it is now an easy (though tedious) task to determinethe size of these sets. For space reasons, we omit the proof and leave the reader with the resulting formulas. Lemma 2.7.
Let m, n ∈ N . Then, | s θ ( m, n ) | = 8( m + 3 m + 4 mn + 2 n + 4 n + 2) | s s θ ( m, n ) | = 8( m + 4 m + 4 mn + 2 n + 6 n + 2) | s s s θ ( m, n ) | = 8( m + 5 m + 4 mn + 2 n + 8 n + 2) | s θ ( m, n ) t m | = 8( m + 4 m + 4 mn + 2 n + 6 n + 3) + 4 | s s θ ( m, n ) t m | = 8( m + 5 m + 4 mn + 2 n + 8 n + 5) | s s s θ ( m, n ) t m | = 8( m + 6 m + 4 mn + 2 n + 10 n + 6) + 4 | s θ ( m, n ) t m s | = 8( m + 5 m + 4 mn + 2 n + 8 n + 4) | s s θ ( m, n ) t m s | = 8( m + 6 m + 4 mn + 2 n + 9 n + 5) + 6 | s s s θ ( m, n ) t m s | = 8( m + 7 m + 4 mn + 2 n + 10 n + 7) + 4 | s θ ( m, n ) t m s t ′ m | = 8( m + 6 m + 4 mn + 2 n + 8 n + 6) + 4 | s s θ ( m, n ) t m s t ′ m | = 8( m + 7 m + 4 mn + 2 n + 9 n + 8) + 4 | s s s θ ( m, n ) t m s t ′ m | = 8( m + 8 m + 4 mn + 2 n + 10 n + 10) + 4 . Remark 2.8.
The big region is made of eight sub-regions, namely, its connected components. In this sectionwe have described just four of them. The remaining four regions are obtained by applying the automor-phism ϕ , and therefore, their description follows from the description of the regions already considered. Inparticular, the size formulas in Corollary 2.5, Lemma 2.6 and Lemma 2.7 are the same for an element andits ϕ -counterpart. In formulas | x | = | x ′ | . In this section we describe some lower intervals for elements located in the thick region. More precisely, weprovide a geometric description of the sets ≤ x n and ≤ e n . We omit description of the lower intervals of theremaining elements in the thick region as this is not needed in the sequel.7e begin by considering the elements x n . In this case we can see there is a pattern that appears modulothree: ≤ x k , ≤ x k +1 and ≤ x k +2 all behave differently. Figure 4 shows some examples. Lemma 2.9.
Let k ≥ . Then ≤ x k = S ( k − , \ Z , where Z is the subset of ∂ ( S ( k − , formed bythe triangles with a side on the line that connects the north and west vertices of S ( k − , .Proof. The result follows by induction on k and the identity ≤ x k +3 = ( ≤ x k )( ≤ t ′ k s t ′ k ) .Having described the set ≤ x k , it is now easy to obtain a geometric description of set ≤ x k +1 . Indeed, ≤ x k +1 = ≤ x k ⊔ ( ∂ t ′ k ( x k )) t ′ k . Similarly, we have ≤ x k +2 = ≤ x k +1 ⊔ ( ∂ s ( x k +1 )) s . Using thesedescriptions and a straightforward counting argument we obtain the following result. Lemma 2.10.
For all k ≥ we have | x k | = 8 k − k, | x k +1 | = 8 k + 4 k, and | x k +2 | = 8 k + 12 k. • (a) From lightest to darkest grey:Lower intervals of x k for k =1 , , and . • (b) From lightest to darkest grey:Lower intervals of x k +1 for k =1 , , and . • (c) From lightest to darkest grey:Lower intervals of x k +2 for k =1 , , and . Figure 4: Lower intervals for elements x n .We now focus on the elements e n . Just as the x n , they obey a pattern that appears modulo . Figure 5shows the different patterns of the lower intervals for the elements e n . In this case it is easier to first considerthe lower intervals of the elements e k +2 . Lemma 2.11.
Let k ≥ . Then the set ≤ e k +2 = S ( k ) where S ( k ) is the smallest square containing e k +2 and x k +2 .Proof. The result follows by induction on k and the identity ≤ e k +5 = ( ≤ e k +2 )( ≤ t k t ′ k s ) .Using Lemma 2.11 as starting point, we can describe the sets ≤ e k +3 and ≤ e k +4 . These are ≤ e k +3 = ( ≤ e k +2 ) ⊔ ( ∂ t k ( ≤ e k +2 )) t k and ≤ e k +4 = ( ≤ e k +3 ) ⊔ ( ∂ t ′ k ( ≤ e k +3 )) t ′ k . An easy counting argument using these descriptions shows the following.
Lemma 2.12.
For all k ≥ , we have | e k | = 8 k + 4 k, | e k +1 | = 8 k + 8 k + 4 , | e k +2 | = 8( k + 1) . We write y ⋖ w to mean that y ≤ w and ℓ ( y ) = ℓ ( w ) − , and ⋖ w to denote the set of all such y . If w = ( s , . . . , s n ) is a reduced expression for w , then declare w ( i ) := s · · · ˆ s i · · · s n = s · · · s i − s i +1 · · · s n .Further, let x n be the preferred reduced expression ( a , · · · , a n ) obtained from the sequence (1.2) and e n =( s , a , a , . . . , a n ) . We also consider coatoms for elements d n defined as the the product of the first n symbolsof the infinite sequence { b i } ∞ i =1 = ( s , s , s , s , s , s , s , s , . . . ) . These elements admit a unique reducedexpression, and they comprise the “northwest wall” of the thin region; we will return to them in Section 6.8 (a) From lightest to darkest grey:Lower intervals of e k +2 for k =0 , , and . • (b) From lightest to darkest grey:Lower intervals of e k +3 for k =0 , , and . • (c) From lightest to darkest grey:Lower intervals of e k +4 for k =0 , , and . Figure 5: Lower intervals for elements e n . Lemma 2.13.
For k ≥ , the sets { y | y ⋖ x n } have the following description. ⋖ x k = { x k (1) , x k (3) , x k (3 k − , x k (3 k ) } = { w k − , s θ ′ ( k − , , θ ( k − , t k − , x k − } ⋖ x k +1 = { x k +1 (1) , x k +1 (3) , x k +1 (3 k ) , x k +1 (3 k + 1) } = { w k − , s θ ′ ( k − , t ′ k − , x k , x k } ⋖ x k +2 = { x k +2 (1) , x k +2 (3) , x k +2 (3 k ) , x k +2 (3 k + 1) , x k +2 (3 k + 2) } = { w k − , s θ ( k − , t ′ k − s , s s s θ ( k − , , θ ( k − , , x k +1 } Furthermore, for k ≥ , ⋖ x k = { w k − , s θ ′ ( k − , t ′ k − s t k − , s s s θ ( k − , t k − , x k − } . Proof.
Recall that all elements y ≤ w can be obtained as subexpressions of a fixed reduced expression for w ,so all elements of ⋖ x n are obtained by removing a single symbol from x n . We will argue for x k , the othercases being similar. We examine which individual symbols can be removed from x k to obtain a reducedexpression. It is clear that x k (1) and x k (3 k ) are reduced, and that none of the s ’s in x k can be removedto obtain a reduced expression. If we remove any a i = s where ≤ i ≤ k − , then there is guaranteed tobe a substring s s s immediately to the right or to the left of a i . Assume it is to the right; upon removal weobtain a substring of the form s s s s s s = s s s s s s = s s s s which is not reduced. An identicalargument applies if the substring is to left, or if a i = s . Thus, the only other possibilities are x k (3) and x k (3 k − , and these are indeed reduced.The latter description of each set ⋖ x n can be easily observed in the geometric realization of W , and theset ⋖ x k is obtained by similar arguments.We will also need the following in Section 5.2. Lemma 2.14.
For k ≥ , the sets ⋖ e n have the following description. ⋖ e k = { x ′ k , x k , s θ ′ ( k − , t ′ k − , e k − } ⋖ e k +1 = { x ′ k +1 , x k +1 , e ′ k , e k } ⋖ e k +2 = { x ′ k +2 , x k +2 , s θ ( k − , , s θ ′ ( k − , , e k +1 } Proof.
Multiplying x n by s on the left does not introduce new possibilities for removal of the symbolsfrom x n , though it can rule them out. We see that removing a from e n gives us a non-reduced substring s s s s s s , but now we can also remove the first symbol s from e n and get x n back, so the size of thecoatom set remains the same. The descriptions above are evident from the geometric realization of W .9 emma 2.15. For n ≥ , the sets { y | y ⋖ d n } have the following description. ⋖ d n = ( { d n (1) , d n (3) , d n ( n − , d n ( n − , d n ( n ) } if n is even , { d n (1) , d n (3) , d n ( n − , d n ( n ) } if n is odd , Proof.
No symbol s or s can be removed from d n and leave a reduced expression unless it is b n . If b i = s for ≤ i ≤ n − , then removal introduces a substring (possibly swapping s with s ) of the form s s s s s s s s = s s s s s s s s = s s s s s s so d n ( i ) is not reduced. The remaining options figure in the lists above.It is clear from Figure 1 that each element in C admits a reduced decomposition of the form x k d n or x k +3 d ′ n . Lemma 2.16.
For any w ∈ C , the set ⋖ w consists of at most five elements. In particular for m, n > wehave ⋖ θ ( m, n ) = { s s θ ′ ( m − , n ) , θ ( m − , n ) t m − s , s s s θ ( m, n − , θ ( m, n − t m s t ′ m } ⋖ θ ( m, n ) t m = { s s θ ′ ( m − , n ) t ′ m − , θ ( m − , n ) t m − s t ′ m − ,s s s θ ( m, n − t m , θ ( m + 1 , n − t m +1 , θ ( m, n ) } ⋖ θ ( m, n ) t m s = { s s θ ′ ( m − , n ) t ′ m − s , θ ( m − , n + 1) ,s s s θ ( m, n − t m s , θ ( m + 1 , n − t m +1 s , θ ( m, n ) t m } ⋖ θ ( m, n ) t m s t ′ m = { s s θ ′ ( m − , n ) t ′ m − s t m − , θ ( m − , n + 1) t m − ,s s s θ ( m, n − t m s t ′ m , θ ( m + 2 , n − , θ ( m, n ) t m s } Proof.
We assume w = x m d n for m = 6 k ; the other case is identical. From Lemma 2.13, x m has fourremovable symbols, but when d n is appended, the last a n can no longer be removed to obtain a reducedexpression. If a = s is removed, then braid relations can be applied to give an expression for x m (3) endingin s which disqualifies this possibility as well. Thus there are only two possibilities for removal from x m : a and a m − .By Lemma 2.15, the symbols of d n give four or five more possibilities depending on the parity of n .However, removal of b , leads to a substring of the form s s s s s s = s s s s . Further, if n = 4 k (respectively, n = 4 k + 2 ), removal of b n − (respectively, b n − ) plus the application of braid relations leadsto an expression for d n ( n − (respectively, d n ( n − ) that begins with s s . As x m ends with s s s ,the result is not a reduced expression. Thus, d m adds three more possibilities for a maximum total of fiveelements.The elements θ ( m, n ) in particular can be written as θ ( m, n ) = x m +1) d n +1 for m odd and θ ( m, n ) = x m +1) d ′ n +1 for m even. An argument similar to the one of the previous paragraph shows that the removalof b n − (or b ′ n − , as the case may be) from this reduced expression yields something not reduced, so ⋖ θ ( m, n ) consists of four elements when m, n > . These are (when m is odd) ⋖ θ ( m, n ) = { x m +1) (1) d n +1 , x m +1) (3 m + 1) d n +1 , x m +1) d m +1 (1) , x m +1) d n +1 (4 n + 1) } . (2.8)The description of this set given in the statement of the lemma can be observed in the geometric realizationof W . The other cases can be reasoned similarly, with all five possibilities yielding reduced expressions. Remark 2.17. If m = 0 , the first two elements of (2.8) are the same, and if n = 0 the last two are thesame, so the set has three elements in case one of m or n is 0. A similar situation occurs for the elements of ⋖ θ ( m, n ) y ( y ∈ { t m , t m s , t m s t ′ m } ) in these cases as well. We leave the precise description to the reader.The following descriptions will be necessary for the arguments of Section 4. They can be deduced byanalysis similar to that of Lemma 2.16. Lemma 2.18.
For all m, n ≥ and y ∈ { t m , t m s , t m s t ′ m } , ⋖ s θ ( m, n ) y = { s } ( ⋖ θ ( m, n ) y ) ∪ { θ ( m, n ) y } ⋖ s s θ ( m, n ) y = { s s } ( ⋖ θ ( m, n ) y ) ∪ { s θ ( m, n ) y } ⋖ s s s θ ( m, n ) y = { s s s } ( ⋖ θ ( m, n ) y ) ∪ { s s θ ( m, n ) y } emark 2.19. Observe that while the description of the set ⋖ θ ( m, n ) y is affected when m or n is zero, therelationship between this set and ⋖ xθ ( m, n ) y is the same in all cases for x ∈ { s , s s , s s s } . In this section we introduce the Hecke algebra of the affine Weyl group of type e B . We also collect severalmultiplicative identities that will be key in order to obtain formulas for Kazhdan-Lusztig basis elements inthe forthcoming sections. e B Let H be the Hecke algebra of W . It is the A := Z [ v, v − ] -algebra with generators H s , H s and H s andrelations H s i = ( v − − v ) H s i + 1 , H s H s = H s H s , H s H s H s H s = H s H s H s H s and H s H s H s H s = H s H s H s H s . Given a reduced expression s i s i . . . s i k of an element w ∈ W we define H w := H s i H s i . . . H s ik . It iswell-known that H w does not depend on the choice of a reduced expression. The set { H w } w ∈ W forms an A -basis of H , which is called the standard basis. It is easy to see that each generator of H is invertible andtherefore all the elements of the standard basis are invertible. There is a Z -linear involution d : H → H which is determined by d ( v ) = v − and d ( H w ) = H − w − . An element invariant under d is called self-dual .There is another basis { H w } w ∈ W called the Kazhdan-Lusztig basis whose elements are uniquely determinedby two conditions: They are self-dual and H w = H w + X x
Given w ∈ W and X ∈ H we denote by G w ( X ) the coefficient of H w in X when it iswritten in terms of the standard basis. That is, X = X w ∈ W G w ( X ) H w . We also define the content of X by c ( X ) := X w ∈ W G w ( X )(1) ∈ Z . Note that this implies that c ( N w ) = | w | and also that c ( X H s ) = 2 c ( X ) for any X ∈ H and any s ∈ S ,since H w H s = ( H ws + v H w , if w < ws ; H ws + v − H w , if w > ws. (3.3)11 efinition 3.2. Let w ∈ W . An element H ∈ H is called triangular of height w if G w ( H ) = 1 and G x ( H ) = 0 for x w . Furthermore, a triangular element H of height w is called monotonic if G x ( H ) ∈ N [ v, v − ] for all x ∈ W and G y ( H ) − v l ( x ) − l ( y ) G x ( H ) ∈ N [ v, v − ] , for all y ≤ x ≤ w .A trivial example of a monotonic element is any N w . We also know that H w is always monotonic (see[BM01, Pla17]). Lemma 3.3.
Let H ∈ H be a monotonic element of height w . Suppose that ws > w . Then, H H s ismonotonic of height ws .Proof. It is clear that H H s is triangular of height ws . On the other hand, (3.3) shows that G x ( H H s ) = (cid:26) vG x ( H ) + G xs ( H ) , if xs > x ; v − G x ( H ) + G xs ( H ) , if xs < x. (3.4)Using this, a simple case analysis shows that the monotonicity of H implies the monotonicity of H H s . Lemma 3.4.
Let ( W, S ) be an arbitrary Coxeter system. Let w ∈ W and s ∈ S such that s ∈ D R ( w ) . Then, N w H s = ( v + v − ) N w .Proof. Since ws < w we can use the Lifting Property [BB06, Proposition 2.2.7] to conclude that multipli-cation on the right by s induces a permutation of the lower interval [ e, w ] . Therefore the result follows by(3.4). Definition 3.5.
Let p and q in A . We write p ≥ q if p − q ∈ N [ v, v − ] . Then for X, Y ∈ H , we write X H ≥ Y if G w ( X ) ≥ G w ( Y ) for all w ∈ W .Notice that X H ≥ Y and c ( X ) = c ( Y ) implies X = Y . We claim no originality in this observation and referto [LP20] for its first application in the context of computation of Kazhdan-Lusztig polynomials. In general,it is not an easy task to prove an inequality of the form X H ≥ Y . However, there are certain situations wherewe can make some simplifications. For instance, let us suppose that Y = X w ∈ Z p w ( v ) N w where Z is a finite subset of W and p z ( v ) = P i ∈ Z p iz v i ∈ N [ v, v − ] . For each i ∈ Z we define Y i = X w ∈ Z p i − ℓ ( w ) w N w . Since the elements Y i lie in different degrees (in the sense that G u ( Y i ) and G u ( Y j ) are monomials of differentdegree if i = j ) in order to prove X H ≥ Y it is enough to show X H ≥ Y i for all i ∈ Z . We stress that Y i = 0 forall but finitely many integers. Henceforth, we refer to this simplification as “degree reasons.”In order to prove inequalities of the form X H ≥ Y i we can make some extra reductions if X is monotonic.For example, if X is monotonic and Y i is made of just one N -element, i.e. Y i = cv k N w , then in order toprove X H ≥ Y i it is enough to check G w ( X ) ≥ cv k . In contrast, if Y i is made of two or more N -elements thenadditional analysis of the relationship between the elements involved in Y i in the Bruhat order is required.We conclude this section with a useful result that allows us to rule out the occurrence of terms in thesum in (3.1). A proof can be found in [KL79, (2.3.f)] or [BB06, Proposition 5.1.9]. Lemma 3.6.
Let x, w ∈ W such that µ ( x, w ) = 0 . Suppose that l ( w ) − l ( x ) > . Then D R ( w ) ⊆ D R ( x ) and D L ( w ) ⊆ D L ( x ) . In particular, if s D R ( w ) and H x appears in the expansion of H w H s when written in terms of the Kazhdan-Lusztig basis, then D R ( w ) ∪{ s } ⊆ D R ( x ) . Similarly, if s D L ( w ) and H x appears in the expansion of H s H w when written in terms of the Kazhdan-Lusztig basis, then D L ( w ) ∪ { s } ⊆ D L ( x ) . .2 Multiplication formulas for N w In this section we collect several multiplicative formulas involving elements N w . Lemma 3.7.
Let m and n be positive integers. Then, N θ ( m,n ) H t m = N θ ( m,n ) t m + v N θ ( m − ,n ) t ′ m . (3.5) Proof.
First, let us prove that the contents on both sides of (3.5) coincide. On the left side, Corollary 2.5shows that c ( N θ ( m,n ) H t m ) = 2 c ( N θ ( m,n ) ) = | θ ( m, n ) | = 16( m + 2 m + 4 mn + 2 n + 2 n + 1) . On the right side, Lemma 2.6 implies that c ( N θ ( m,n ) t m + v N θ ( m − ,n ) t ′ m ) = | θ ( m, n ) t m | + | θ ( m − , n ) t ′ m | = 16( m + 2 m + 4 mn + 2 n + 2 n + 1) . Thus we only need to show that N θ ( m,n ) H t m H ≥ N θ ( m,n ) t m + v N θ ( m − ,n ) t ′ m . By degree reasons it is enough to check N θ ( m,n ) H t m H ≥ N θ ( m,n ) t m , (3.6) N θ ( m,n ) H t m H ≥ v N θ ( m − ,n ) t ′ m . (3.7)Furthermore, by the monotonicity of N θ ( m,n ) H t m ensured by Lemma 3.3, inequalities (3.6) and (3.7) willfollow from the inequalities G θ ( m,n ) t m ( N θ ( m,n ) H t m ) ≥ , (3.8) G θ ( m − ,n ) t ′ m ( N θ ( m,n ) H t m ) ≥ v, (3.9)respectively. Inequality (3.8) is clear. On the other hand, a direct computation shows that G θ ( m − ,n ) t ′ m ( v H θ ( m − ,n ) t ′ m H t m ) = v, which proves (3.9). The lemma is proved. Remark 3.8.
Henceforth, we adopt the convention that N xθ ( m,n ) y and H xθ ( m,n ) y are equal to zero whenever m or n is negative, for any x, y ∈ W . This will allow us to express some results more compactly andconsistently.Using the same arguments as in the proof of Lemma 3.7 we obtain the following four lemmas. Lemma 3.9.
Let m and n be positive integers. We have N θ (0 ,n ) H s = N θ (0 ,n ) s + v N θ (0 ,n − s ; N θ ( m, H t m = N θ ( m, t m + v N θ ( m − , t ′ m . Furthermore, N θ (0 , H s = N θ (0 , s . Lemma 3.10.
For all m ∈ N we have H s N θ ( m, = N s θ ( m, + v N s θ ′ ( m − , , where θ ′ ( m − , denotes the image of θ ( m − , under the automorphism ϕ . Lemma 3.11.
Let m be a positive integer. We have N s θ ( m, H t m = N s θ ( m, t m + v N s θ ( m − , t ′ m . Furthermore, N s θ (0 , H s = N s θ (0 , s + v N s s . emma 3.12. Let m be a positive integer. We have N s s s θ ( m, H t m = N s s s θ ( m, t m + v N s s s θ ( m − , t ′ m . The following lemma is proved using essentially the same argument as the one used in the proof of Lemma3.7. This, however, is the first time so far in which we have a case where we must show X H ≥ Y with Y consisting of two or more N -terms, which leads some extra subtleties in the analysis. Lemma 3.13.
Let m and n be positive integers. Then, N θ ( m,n ) H t m H s H t m = 2 N θ ( m,n ) H t m + N θ ( m +1 ,n ) + N θ ( m − ,n +1) + N θ ( m +1 ,n − + N θ ( m − ,n ) . (3.10) Proof.
Let us denote by L and R the left-hand and right-hand side of (3.10), respectively. A repeatedapplication of Corollary 2.5 shows that c ( L ) = c ( R ) = 64( m + 2 m + 4 mn + 2 n + 2 n + 1) . Therefore, to finish the proof we only need to show that L H ≥ R . Lemma 3.7 implies R = 2 N θ ( m,n ) t m + 2 v N θ ( m − ,n ) t ′ m + N θ ( m +1 ,n ) + N θ ( m − ,n +1) + N θ ( m +1 ,n − + N θ ( m − ,n ) . Hence, by degree reasons, it is enough to check L H ≥ N θ ( m +1 ,n ) (3.11) L H ≥ N θ ( m − ,n ) (3.12) L H ≥ N θ ( m − ,n +1) + 2 N θ ( m,n ) t m (3.13) L H ≥ N θ ( m +1 ,n − + 2 v N θ ( m − ,n ) t ′ m (3.14)We now notice that L is monotonic of height θ ( m + 1 , n ) by Lemma 3.3. It is clear that G θ ( m +1 ,n ) ( L ) =1 . Then the monotonicity of L shows (3.11). On the other hand, a direct computation reveals that G θ ( m − ,n ) ( v H θ ( m − ,n ) H t m H s H t m ) = v + 1 . Therefore, G θ ( m − ,n ) ( L ) ≥ G θ ( m − ,n ) ( v H θ ( m − ,n ) H t m H s H t m ) = v + 1 ≥ . The above inequality together with the monotonicity of L proves (3.12).We now prove (3.13). Since we have two N -terms that cannot be separated by degree reasons on theright-hand side, we need to proceed in a different way. Let us explain this more generally. Suppose we wantto prove L H ≥ c v i N x + c v j N y , (3.15)for some positive integers c and c , and where j − i = l ( x ) − l ( y ) (otherwise, degree reasons allow us to splitthe inequality in (3.15) in two inequalities with only one N -term on the right-hand side). Further supposethat ≤ x ∩ ≤ y = ≤ z ∪ ≤ w for some elements z and w . Then, by the monotonicity of L , (3.15) will followfrom L H ≥ c v i N x ,L H ≥ c v j N y ,L H ≥ ( c + c ) v i + l ( x ) − l ( z ) N z , and L H ≥ ( c + c ) v i + l ( x ) − l ( w ) N w . (3.16)Let us apply this general principle to prove (3.13). By the diagrammatic description of lower intervals givenin §2.1 we have ( ≤ θ ( m − , n + 1)) ∩ ( ≤ θ ( m, n ) t m ) = ( ≤ θ ( m − , n ) t ′ m s t m ) . Therefore, we only need to prove G θ ( m − ,n +1) ( L ) ≥ , G θ ( m,n ) t m ( L ) ≥ and G θ ( m − ,n ) t ′ m s t m ( L ) ≥ v. G θ ( m − ,n +1) ( v H θ ( m − ,n ) t ′ m s H t m H s H t m ) = 1 G θ ( m,n ) t m ( H θ ( m,n ) H t m H s H t m ) = 2 G θ ( m − ,n ) t ′ m s t m ( v H θ ( m − ,n ) t ′ m H t m H s H t m ) = vG θ ( m − ,n ) t ′ m s t m ( v H θ ( m − ,n ) t ′ m s H t m H s H t m ) = 2 v. The proof of (3.14) is similar, noting that ( ≤ θ ( m + 1 , n − ∩ ( ≤ θ ( m − , n ) t ′ m ) = ( ≤ θ ( m − , n )) ∪ ( ≤ θ ( m, n − t m ) . This description of the lower intervals allows one to apply the discussion leading up to (3.16); we leave thedetails to the reader.
Proposition 3.14.
Let m and n be positive integers. Then, N θ ( m,n ) ( H t m H s H t m − H t m ) = N θ ( m +1 ,n ) + N θ ( m − ,n +1) + N θ ( m +1 ,n − + N θ ( m − ,n ) . Proof.
The result is an immediate consequence of Lemma 3.13.Similarly, we have the following proposition.
Proposition 3.15.
Let m, n ∈ N . We have N θ ( m, ( H t m H s H t m − H t m ) = N θ ( m +1 , + N θ ( m − , + (1 + v ) N θ ( m − , , N θ (0 ,n ) ( H s H s H s − H s ) = N θ (1 ,n ) + (1 + v ) N θ (1 ,n − + v N θ (1 ,n − . Propositions 3.14 and 3.15 should be understood as a way to pass from N θ ( m,n ) to N θ ( m +1 ,n ) . We nowfocus on obtaining a similar statement but this time we want to pass from N θ (0 ,n ) to N θ (0 ,n +1) . Lemma 3.16.
Let n be a positive integer. We have N θ (0 ,n ) s H s + v N θ (0 ,n − s s = N θ (0 ,n ) s s + N θ (0 ,n ) + v N θ (1 ,n − . (3.17) Proof.
Once we notice that the element on the left-hand side of (3.17) is monotonic (of height θ (0 , n ) s s )we can argue as in the proof of Lemma 3.7. Lemma 3.17.
Let n be an integer greater than . We have N θ (0 ,n +1) N θ (0 ,n ) s s + v N θ (0 ,n ) + N θ (2 ,n − + N θ (0 ,n ) H s H s H s H s + 2 v N θ (0 ,n − s s = 3 N θ (0 ,n ) + 2 v N θ (1 ,n − + v N θ (2 ,n − + v − N θ (0 ,n ) + 3 v N θ (0 ,n − + 2 v N θ (1 ,n − + N θ (0 ,n − . (3.18) Proof.
Let L and R be the left and right sides of (3.18), respectively. Corollary 2.5 and Lemma 2.6 imply c ( L ) = c ( R ) = 8(36 n + 26 n + 18) . Therefore, in order to prove L = R we only need to show that L H ≥ R . By degree reasons, it is enough toprove the following inequalities L H ≥ N θ (0 ,n +1) , (3.19) L H ≥ N θ (0 ,n ) s s + v N θ (0 ,n ) + N θ (2 ,n − , (3.20) L H ≥ N θ (0 ,n ) + 2 v N θ (1 ,n − + v N θ (2 ,n − , (3.21) L H ≥ v − N θ (0 ,n ) + 3 v N θ (0 ,n − + 2 v N θ (1 ,n − , (3.22) L H ≥ N θ (0 ,n − . (3.23)15sing Lemma 3.3 we can see that N θ (0 ,n ) H s H s H s H s is a monotonic element of height θ (0 , n + 1) .Furthermore, N θ (0 ,n − s s is clearly monotonic. Since θ (0 , n − s s ≤ θ (0 , n + 1) we conclude that L ismonotonic.An easy computation shows that G θ (0 ,n +1) ( H θ (0 ,n ) H s H s H s H s ) = 1 ,G θ (0 ,n − ( v H θ (0 ,n − s H s H s H s H s ) = 1 + v . This allows us to conclude that G θ (0 ,n +1) ( L ) ≥ and G θ (0 ,n − ( L ) ≥ . Then, the monotonicity of L proves(3.19) and (3.23).We now focus on (3.20). We notice that θ (0 , n ) is smaller than θ (0 , n ) s s and θ (2 , n − , but the latterelements are incomparable in Bruhat order. Furthermore, by the description of lower intervals given in §2.1we have ( ≤ θ (0 , n ) s s ) ∩ ( ≤ θ (2 , n − ≤ θ (1 , n − s s ) . Finally, we notice that θ (0 , n ) ≤ θ (1 , n − s s . Thus, by the monotonicity of L , in order to prove (3.20) itis enough to show that G θ (0 ,n ) s s ( L ) ≥ (3.24) G θ (2 ,n − ( L ) ≥ (3.25) G θ (1 ,n − s s ( L ) ≥ v ; (3.26) G θ (0 ,n ) ( L ) ≥ v . (3.27)A direct computation shows that G θ (0 ,n ) s s (cid:0) H θ (0 ,n ) H s H s H s H s (cid:1) = 2 . This proves (3.24). Similarly, wehave G θ (2 ,n − (cid:0) v H θ (0 ,n − s s s H s H s H s H s (cid:1) = 1 , which proves (3.25). On the other hand, we have G θ (1 ,n − s s ( v H θ (0 ,n − s s s H s H s H s H s ) = 2 v ; G θ (1 ,n − s s ( v H θ (0 ,n − s s H s H s H s H s ) = v. This proves (3.26). Finally, we have G θ (0 ,n ) (cid:0) v H θ (0 ,n − s s s H s H s H s H s (cid:1) = v ; G θ (0 ,n ) (cid:0) v H θ (0 ,n − s s H s H s H s H s (cid:1) = v ; G θ (0 ,n ) (cid:0) v H θ (0 ,n − s H s H s H s H s (cid:1) = v ; G θ (0 ,n ) (cid:0) v H d n +2 H s H s H s H s (cid:1) = v . These four equalities show (3.27). This completes the proof of inequality (3.20). The remaining two inequal-ities (3.21) and (3.22) are treated similarly.We now are in a position to obtain a multiplication formula that relates N θ (0 ,n ) and N θ (0 ,n +1) . Proposition 3.18.
Let n be an integer greater than . Then, N θ (0 ,n ) ( H s H s H s H s − H s H s + 1 − ( v + v − ) ) = N θ (0 ,n +1) + (1 + v ) N θ (0 ,n − + v N θ (2 ,n − + N θ (2 ,n − . Proof.
By combining Lemma 3.9 and Lemma 3.16 we obtain N θ (0 ,n ) H s H s = N θ (0 ,n ) s s + N θ (0 ,n ) + v N θ (1 ,n − + v N θ (0 ,n − + v N θ (1 ,n − − v N θ (0 ,n − s s . (3.28)Hence, the result follows by combining Lemma 3.17 and (3.28).So far we have proved several multiplicative identities involving N -elements. These will be sufficientto obtain formulas for Kazhdan-Lusztig basis elements corresponding to elements in the big region in §4.We end this section by stating some extra multiplicative identities that will be useful to prove formulas forKazhdan-Lusztig basis elements corresponding to elements in the thick region in §5.16 emma 3.19. For m ≥ , the element (cid:0) H s H s H s − H s (cid:1) N θ ( m, is equal to N s s s θ ( m, + ( v − + 2 v ) N θ ( m, + v ( N θ ( m − , + N s θ ′ ( m − , ) + v ( N s θ ( m − , − N s s s θ ( m − , ) . (3.29) Furthermore, we have (cid:0) H s H s H s − H s (cid:1) N θ (0 , = N s s s θ (0 , + ( v + v − ) N θ (0 , . (3.30) Proof.
Equation (3.30) can be verified manually. Similarly, we can check (3.29) for m = 1 . We stress thatin this case some terms vanish in accordance with Remark 3.8. From now and on we fix m ≥ . We define L := H s H s H s N θ ( m, + v N s s s θ ( m − , and R := H s N θ ( m, + N s s s θ ( m, + ( v − + 2 v ) N θ ( m, + v ( N θ ( m − , + N s θ ′ ( m − , ) + v N s θ ( m − , . It is clear that the claim in the lemma is equivalent to L = R , which we will now prove. Using Lemma 3.10,we can rewrite R as R = N s θ ( m, + N s s s θ ( m, + ( v − + 2 v ) N θ ( m, + v ( N θ ( m − , + 2 N s θ ′ ( m − , ) + v N s θ ( m − , . A straightforward computation using Corollary 2.5 and Lemma 2.7 shows c ( L ) = c ( R ) = 8(9 m + 17 m + 4) . To complete the proof we only need to show that L H ≥ R . By degree reasons, this inequality will follow fromthe next four inequalities L H ≥ N s s s θ ( m, (3.31) L H ≥ N s θ ( m, + 2 v N θ ( m, (3.32) L H ≥ v − N θ ( m, + 2 v N s θ ′ ( m − , + v N θ ( m − , (3.33) L H ≥ v N s θ ( m − , . (3.34)We notice that the left version of Lemma 3.3 shows that L is monotonic. Inequalities (3.31), (3.32) and(3.34) are easily obtained, so we will focus on (3.33). We notice that θ ( m, is greater than s θ ′ ( m − , and than θ ( m − , but the last two elements are incomparable in Bruhat order. By the diagrammaticdescription of the lower intervals given in §2.1 we obtain ( ≤ s θ ′ ( m − , ∩ ( ≤ θ ( m − , ≤ s s s θ ( m − , . Therefore, by the monotonicity of L , (3.33) reduces to proving the following polynomial inequalities G θ ( m, ( L ) ≥ v − , (3.35) G s θ ′ ( m − , ( L ) ≥ v, (3.36) G θ ( m − , ( L ) ≥ v, (3.37) G s s s θ ( m − , ( L ) ≥ v . (3.38)Set X := H s H s H s . A direct computation (using the left version of (3.3)) shows that G θ ( m, ( X H θ ( m, ) = v − , proving (3.35). Similarly, we have G s θ ′ ( m − , ( X ( v H s s θ ′ ( m − , )) = G s θ ′ ( m − , ( X ( v H s θ ′ ( m − , )) = G s θ ′ ( m − , ( X ( v H θ ′ ( m − , )) = v. These three equalities together prove (3.36). We now observe that G θ ( m − , ( X ( v H s θ ′ ( m − , )) = G θ ( m − , ( X ( v H θ ( m − , )) = v. This proves (3.37). Finally, if we set z := s s s θ ( m − , then we have G z ( X ( v H s θ ′ ( m − , )) = G z ( X ( v H θ ( m − , )) = G z ( X ( v H θ ′ ( m − , )) = v . These three equalities show that G z ( X N θ ( m, ) ≥ v . Since G z ( v N z ) = v , inequality (3.38) follows fromthe definition of L . The lemma is proved. 17sing similar arguments we can prove the following result, recalling the definitions of x n , e n and u n fromthe introduction. The details are left to the reader. Lemma 3.20.
Given an integer k we set f ( k ) = 3 k + 1 . Let Y = H s H s H s − H s . For k ≥ , N x f ( k ) Y = N x f ( k +1) + N x f ( k − + N θ ( k − , t k − + N s s s θ ( k − , t k − + N s θ ′ ( k − , t ′ k − + v N θ ( k − , t k − N u f ( k ) Y = N u f ( k +1) + N u f ( k − + N θ ( k − , t k − + ( v + v − ) N θ ( k − , t k − + v N u f ( k − N e f ( k ) Y = N e f ( k +1) + N e f ( k − + N s θ ( k − , t k − + N s θ ′ ( k − , t ′ k − + v N s θ ( k − , t k − + v N s θ ′ ( k − , t ′ k − . In each case, the identity also holds if Y is replaced by Y ′ = H s H s H s − H s . Remark 3.21.
The reader may have noticed that in the second formula N u f ( k − shows up twice withdifferent coefficients. This may appear to be a slip, but it is not. The reason behind this choice will becomeclear in Section 5. In this section we provide an explicit formula for Kazhdan-Lusztig basis elements indexed by elements locatedin the big region. C Let us recall a definition from the introduction.
Definition 4.1.
Let ( m, n ) ∈ N . We define Supp( m, n ) = (cid:26) { ( m − i, n − j ) ∈ N | i, j ∈ N } , if m is odd; { ( m − i, n − j ) ∈ N | i, j ∈ N } − { (0 , b ) | b n mod 2 } , if m is even.We also define ˆ H θ ( m,n ) := X ( a,b ) ∈ Supp( m,n ) v ( m − a )+2( n − b ) N θ ( a,b ) . The first goal in this section is to prove that H θ ( m,n ) = ˆ H θ ( m,n ) for all ( m, n ) ∈ N . To do this we needseveral preliminary lemmas. Lemma 4.2.
Let m and n be positive integers. Then, ˆ H θ ( m,n ) = N θ ( m,n ) + v ˆ H θ ( m − ,n ) + v ˆ H θ ( m,n − − v ˆ H θ ( m − ,n − . Furthermore, ˆ H θ (0 ,n ) = N θ (0 ,n ) + v ˆ H θ (0 ,n − and ˆ H θ ( m, = N θ ( m, + v ˆ H θ ( m − , . (4.1) Proof.
The result follows directly from the definition of the elements ˆ H θ ( m,n ) once we recall our conventionabout negative indices in Remark 3.8. Lemma 4.3.
Let ( m, n ) ∈ N . Then ˆ H θ ( m,n ) ( H t m H s H t m − H t m ) = ˆ H θ ( m +1 ,n ) + ˆ H θ ( m − ,n +1) + ˆ H θ ( m +1 ,n − + ˆ H θ ( m − ,n ) . (4.2) Proof.
A direct computation shows that (4.2) holds for the pairs (0 , , (1 , , (0 , , (2 , , (1 , and (0 , .We now fix u ≥ and assume that (4.2) holds for all pairs ( m ′ , n ′ ) with m ′ + n ′ < u . We fix a pair ( m, n ) with m + n = u . We split the proof in four cases. We recall that the set of D R ( θ ( a, b )) only depends on theparity of a . For the sake of brevity we write X ↑ := H t m H s H t m − H t m . ase A. Suppose that m ≥ and n ≥ .Using Proposition 3.14, Lemma 4.2 and our inductive hypothesis we get ˆ H θ ( m,n ) X ↑ = N θ ( m +1 ,n ) + N θ ( m − ,n +1) + N θ ( m +1 ,n − + N θ ( m − ,n ) + v ˆ H θ ( m − ,n ) + v ˆ H θ ( m − ,n +1) + v ˆ H θ ( m − ,n − + v ˆ H θ ( m − ,n ) + v ˆ H θ ( m +1 ,n − + v ˆ H θ ( m − ,n ) + v ˆ H θ ( m +1 ,n − + v ˆ H θ ( m − ,n − − v ˆ H θ ( m − ,n − − v ˆ H θ ( m − ,n ) − v ˆ H θ ( m − ,n − − v ˆ H θ ( m − ,n − . (4.3)Then, by adding by columns in the right-hand side of (4.3) and using Lemma 4.2 we obtain (4.2). Case B.
Suppose that m = 1 and n ≥ .The main difference with respect to the above Case A is that when we decompose ˆ H θ (1 ,n ) using Lemma4.2 there are two terms that do not appear. More concretely, we have ˆ H θ (1 ,n ) = N θ (1 ,n ) + v ˆ H θ (1 ,n − . Using Proposition 3.14 and our inductive hypothesis we get ˆ H θ (1 ,n ) X ↑ = N θ (2 ,n ) + N θ (0 ,n +1) + N θ (2 ,n − + N θ (0 ,n ) + v ˆ H θ (2 ,n − + v ˆ H θ (0 ,n ) + v ˆ H θ (2 ,n − + v ˆ H θ (0 ,n − . (4.4)We now use (4.1) to rewrite N θ (0 ,n +1) and N θ (0 ,n ) as N θ (0 ,n +1) = ˆ H θ (0 ,n +1) − v ˆ H θ (0 ,n − and N θ (0 ,n ) = ˆ H θ (0 ,n ) − v ˆ H θ (0 ,n − within (4.4). Another application of Lemma 4.2 gives us (4.2). Case C.
Suppose that m = 0 and n ≥ .As in the previous cases we use Lemma 4.2 in order to get ˆ H θ (0 ,n ) = N θ (0 ,n ) + v ˆ H θ (0 ,n − . Using Proposition 3.15 and our inductive hypothesis we get ˆ H θ (0 ,n ) X ↑ = ( N θ (1 ,n ) + v N θ (1 ,n − + v ˆ H θ (1 ,n − ) + ( N θ (1 ,n − + v N θ (1 ,n − + v ˆ H θ (1 ,n − ) . Finally, two applications of Lemma 4.2 give (4.2).
Case D. m ≥ and n = 0 .By combining (4.1), Proposition 3.15 and our inductive hypothesis we get ˆ H θ ( m, X ↑ = ( N θ ( m +1 , + v ˆ H θ ( m − , ) + ( N θ ( m − , + v N θ ( m − , + v ˆ H θ ( m − , ) + ( N θ ( m − , + v ˆ H θ ( m − , )= ˆ H θ ( m +1 , + ˆ H θ ( m − , + ˆ H θ ( m − , . This completes the proof of the lemma.
Lemma 4.4.
For all n ∈ N we have ˆ H θ (0 ,n ) ( H s H s H s H s − H s H s + 1 − ( v + v − ) ) = ˆ H θ (0 ,n +1) + ˆ H θ (2 ,n − + ˆ H θ (0 ,n − . (4.5) Proof.
We proceed by induction on n . The result can be checked directly for n ≤ , so we assume that n > and that (4.5) holds for all n ′ < n . We write X ↑ := H s H s H s H s − H s H s + 1 − ( v + v − ) . By combining Proposition 3.18, Lemma 4.2 and our inductive hypothesis we obtain (cid:16) N θ (0 ,n +1) + v ˆ H θ (0 ,n − (cid:17) +ˆ H θ (0 ,n ) X ↑ = (cid:16) N θ (2 ,n − + v N θ (2 ,n − + v N θ (0 ,n − + v ˆ H θ (2 ,n − (cid:17) + (cid:16) N θ (0 ,n − + v ˆ H θ (0 ,n − (cid:17) = ˆ H θ (0 ,n +1) + ˆ H θ (2 ,n − + ˆ H θ (0 ,n − , ˆ H θ (2 ,n − = N θ (2 ,n − + v N θ (2 ,n − + v N θ (0 ,n − + v ˆ H θ (2 ,n − , which follows directly from the definition of ˆ H θ (2 ,n − . Theorem 4.5.
For all ( m, n ) ∈ N we have ˆ H θ ( m,n ) = H θ ( m,n ) .Proof. By definition of ˆ H θ ( m,n ) it is clear that G θ ( m,n ) ( ˆ H θ ( m,n ) ) = 1 and that G x ( ˆ H θ ( m,n ) ) ∈ v N [ v ] for all x < θ ( m, n ) . Therefore, in order to prove the theorem it is enough to show that ˆ H θ ( m,n ) is self-dual. To dothis we proceed by induction on m + n . The result is clear for the pair (0 , since H θ (0 , = N θ (0 , = ˆ H θ (0 , .We fix u ∈ N and assume that ˆ H θ ( m,n ) is self-dual for all pairs ( m, n ) such that m + n ≤ u . By Lemma 4.3we know that ˆ H θ ( m,n ) ( H t m H s H t m − H t m ) = ˆ H θ ( m +1 ,n ) + ˆ H θ ( m − ,n +1) + ˆ H θ ( m +1 ,n − + ˆ H θ ( m − ,n ) . We notice that H t m H s H t m − H t m is self-dual. Also, the terms ˆ H θ ( m,n ) , ˆ H θ ( m − ,n +1) , ˆ H θ ( m +1 ,n − and ˆ H θ ( m − ,n ) are self-dual by our inductive hypothesis. We can thus conclude that ˆ H θ ( m +1 ,n ) is self-dual aswell. This shows that ˆ H θ ( a,b ) is self-dual for all the pairs ( a, b ) with a + b = u + 1 and a = 0 .It remains to show that ˆ H θ (0 ,u +1) is self-dual. By Lemma 4.4 we know that ˆ H θ (0 ,u ) ( H s H s H s H s − H s H s + 1 − ( v + v − ) ) = ˆ H θ (0 ,u +1) + ˆ H θ (2 ,u − + ˆ H θ (0 ,u − . We notice that H s H s H s H s − H s H s + 1 − ( v + v − ) is self-dual and that ˆ H θ (0 ,u − is also self-dual byour inductive hypothesis. Although the self-duality of ˆ H θ (2 ,u − is not covered by our inductive hypothesis,we have already proved its self-duality since (2 , u − is a pair ( a, b ) with a = 0 . As before, we concludethat ˆ H θ (0 ,u +1) is self-dual as well, and the theorem is proved. Remark 4.6.
There is an alternative way to prove Theorem 4.5 using the results in [LPP21]. We prefer topresent the proof just given in order to keep this paper self-contained and cite only available sources.
Theorem 4.7.
Let ( m, n ) ∈ N . Then, H θ ( m,n ) H t m = H θ ( m,n ) t m (4.6) H θ ( m,n ) t m H s = H θ ( m,n ) t m s + H θ ( m,n ) (4.7) H θ ( m,n ) t m s H t ′ m = H θ ( m,n ) t m s t ′ m + H θ ( m,n ) t m . (4.8) Proof.
By Lemma 2.16, we have xt m > x for all x ⋖ θ ( m, n ) . On the other hand, Theorem 4.5 implies that µ ( x, θ ( m, n )) = 0 if and only if x ⋖ θ ( m, n ) . Hence, (3.1) implies (4.6).In order to prove (4.7) we begin by noticing that Lemma 2.16 implies that the only element in ⋖ θ ( m, n ) t m satisfying xs < x is θ ( m, n ) . Therefore, a combination of (3.1) and Lemma 3.6 yields H θ ( m,n ) t m H s = H θ ( m,n ) t m s + H θ ( m,n ) + X y ∈ WD R ( y )= { t m ,t ′ m ,s } m y H y , (4.9)where m y ∈ N . We stress that D R ( θ ( m, n ) t m ) = { t m , t ′ m } . Since there is no y ∈ W with D R ( y ) = { t m , t ′ m , s } we conclude that the sum in (4.9) is empty. This gives (4.7).We now prove (4.8). Once again, Lemma 2.16 shows that the only element in ⋖ θ ( m, n ) t m s satisfying xt ′ m < x is θ ( m, n ) t m . Hence (3.1) and Lemma 3.6 give H θ ( m,n ) t m s H t ′ m = H θ ( m,n ) t m s t ′ m + H θ ( m,n ) t m + X y ∈ Y n y H y , (4.10)where n y ∈ N and Y = { y ∈ W | D R ( y ) = { t ′ m , s } , D L ( y ) = { s , s }} .By the discussion at the beginning of §2.1 we conclude that y ∈ Y if and only if there exist ( a, b ) ∈ N with a ≡ m mod 2 such that y = θ ( a, b ) . We now multiply (4.10) on the right by H t m and using (3.2) and(4.6) we obtain H θ ( m,n ) t m s H t ′ m H t m = H θ ( m,n ) t m s t ′ m H t m + ( v + v − ) H θ ( m,n ) t m + X y ∈ Y n y H yt m . (4.11)20n the other hand, we combine (3.2), Lemma 4.3, Theorem 4.5, (4.6) and (4.7) to obtain H θ ( m,n ) t m s H t m H t ′ m = ( v + v − ) H θ ( m,n ) t m + H θ ( m +1 ,n ) t ′ m + H θ ( m − ,n +1) t ′ m + H θ ( m +1 ,n − t ′ m + H θ ( m − ,n ) t ′ m . (4.12)We now notice that H t m H t ′ m = H t ′ m H t m and conclude that the right-hand side of (4.11) and (4.12) coincide.Therefore, after cancelling out the term ( v + v − ) H θ ( m,n ) t m we get H θ ( m,n ) t m s t ′ m H t m + X y ∈ Y n y H yt m = H θ ( m +1 ,n ) t ′ m + H θ ( m − ,n +1) t ′ m + H θ ( m +1 ,n − t ′ m + H θ ( m − ,n ) t ′ m . (4.13)Suppose that n y = 0 for some y = θ ( a, b ) ∈ Y . Since a ≡ m mod 2 we know that a = m ± . In particular, θ ( a, b ) t m is not equal to any of the elements that index the Kazhdan-Lusztig basis elements on the right-handside of (4.13). We reach a contradiction and conclude that n y = 0 for all y ∈ Y . If we look back at (4.10)then we see that it reduces to (4.8). Let us remark that Theorem 4.5 and Theorem 4.7 give us all the Kazhdan-Lusztig basis elements indexed byelements in the fundamental region C and also in the region ϕ ( C ) . In this section we extend this result to allthe elements in the big region. We stress that we only need specify the description of the Kazhdan-Lusztigbasis elements for the regions s C , s s C and s s s C since the description for the elements in the regions s ϕ ( C ) , s s ϕ ( C ) and s s s ϕ ( C ) will follow from the above by applying the automorphism ϕ . Theorem 4.8.
Let ( m, n ) ∈ N . Then, H s H θ ( m,n ) = H s θ ( m,n ) (4.14) H s H θ ( m,n ) t m = H s θ ( m,n ) t m (4.15) H s H θ ( m,n ) t m s = H s θ ( m,n ) t m s (4.16) H s H θ ( m,n ) t m s t ′ m = H s θ ( m,n ) t m s t ′ m (4.17) H s H s θ ( m,n ) = H s s θ ( m,n ) + H θ ( m,n ) (4.18) H s H s θ ( m,n ) t m = H s s θ ( m,n ) t m + H θ ( m,n ) t m (4.19) H s H s θ ( m,n ) t m s = H s s θ ( m,n ) t m s + H θ ( m,n ) t m s (4.20) H s H s θ ( m,n ) t m s t ′ m = H s s θ ( m,n ) t m s t ′ m + H θ ( m,n ) t m s t ′ m (4.21) H s H s s θ ( m,n ) = H s s s θ ( m,n ) + H s θ ( m,n ) (4.22) H s H s s θ ( m,n ) t m = H s s s θ ( m,n ) t m + H s θ ( m,n ) t m (4.23) H s H s s θ ( m,n ) t m s = H s s s θ ( m,n ) t m s + H s θ ( m,n ) t m s (4.24) H s H s s θ ( m,n ) t m s t ′ m = H s s s θ ( m,n ) t m s t ′ m + H s θ ( m,n ) t m s t ′ m . (4.25) Proof.
We fix ( m, n ) ∈ N and let R := { e, t m , t m s , t m s t ′ m } . By our description of the set of coatomsfor lower intervals in Lemma 2.16 we have s x > x for all y ∈ R and x ⋖ θ ( m, n ) y . Therefore, since D L ( θ ( m, n ) y ) = { s , s } for all y ∈ R and there is no element z ∈ W with D L ( z ) = { s , s , s } , we obtainvia the left version of (3.1) and Lemma 3.6 all the identities (4.14)–(4.17). We pause our proof for a momentto make an observation that will be useful for the rest of the proof. Remark 4.9.
We remark that (4.14)–(4.17) together are equivalent to the following: For any z ∈ C we have H s H z = H s z . Of course, by applying ϕ to the above equality we get H s H u = H s u for all u ∈ ϕ ( C ) .We continue with the proof of the next four identities. Now, Lemma 2.18 allows us to conclude that foreach y ∈ R the only element in ⋖ s θ ( m, n ) y satisfying s x < x is θ ( m, n ) y . Since D L ( s θ ( m, n ) y ) = { s , s } for all y ∈ R , another application of the left version of (3.1) and Lemma 3.6 give us (4.18)–(4.21).For the last four identities we need to work a little harder. In this case we have D L ( s s θ ( m, n ) y ) = { s } for all y ∈ R . Therefore, the left version of (3.1) and Lemma 3.6 only allow us to conclude that for each y ∈ R we have H s H s s θ ( m,n ) y = H s s s θ ( m,n ) y + H s θ ( m,n ) y + X z ∈ C m yz H z , (4.26)21here m yz ∈ N .We now multiply (4.26) by H s on the left, and using (3.2) together with Remark 4.9 we get H s H s H s s θ ( m,n ) y = H s H s s s θ ( m,n ) y + ( v + v − ) H s θ ( m,n ) y + X z ∈ C m yz H s z . (4.27)We now compute H s H s H s s θ ( m,n ) y in a different way using the fact that H s H s = H s H s . Lemma 2.18shows that for all y ∈ R the only element in ⋖ s s θ ( m, n ) y satisfying s x < x is s θ ( m, n ) y . On the otherhand, we notice that s s s θ ( m, n ) y belongs to ϕ ( C ) since D L ( s s s θ ( m, n ) y ) = { s , s } . Using these twofacts together with the left version of (3.1) and Lemma 3.6 we obtain H s H s H s s θ ( m,n ) y = H s H s s s θ ( m,n ) y + H s θ ( m,n ) y + X u ∈ ϕ ( C ) n yu H u = H s s s s θ ( m,n ) y + ( v + v − ) H s θ ( m,n ) y + X u ∈ ϕ ( C ) n yu H s u , (4.28)for some n yu ∈ N .We now fix y ∈ R and suppose that m yz = 0 for some z ∈ C . By comparing the right-hand sides of (4.27)and (4.28), we conclude that s z ∈ s ϕ ( C ) , which is absurd since s C ∩ s ϕ ( C ) = ∅ . Therefore, m yz = 0 forall z ∈ C and (4.26) reduces to H s H s s θ ( m,n ) y = H s s s θ ( m,n ) y + H s θ ( m,n ) y . Since this identity holds for all y ∈ R we obtain (4.22)–(4.25). Remark 4.10.
The formulas in Theorem 4.7 and Theorem 4.8 are apparently different from the onespresented in Theorem 1.1. It is an easy exercise (which is left to the reader) to show that both are equivalent.
Formulas given in §4.1 and §4.2 enable the efficient computation of any Kazhdan-Lusztig polynomial h x,w ( v ) for all x ∈ W and all w in the big region. In order to obtain formulas for the thick region in §5, it will beconvenient to have more explicit expressions on hand for certain Kazhdan-Lusztig basis elements in the bigregion. This is the main goal of this section. Lemma 4.11.
Let m ≥ . We have H θ ( m, t m = N θ ( m, t m + v H θ ( m − , t m − .Proof. We proceed by induction on m . The cases m = 0 and m = 1 are easily checked. We assume m > and that the lemma holds for m − . Using Lemma 3.9, the explicit formula for H θ ( m, , (4.6), our inductivehypothesis and the fact that t m = t m − we obtain H θ ( m, t m = H θ ( m, H t m = ( N θ ( m, + v H θ ( m − , ) H t m = N θ ( m, t m + v N θ ( m − , t m − + v H θ ( m − , t m = N θ ( m, t m + v ( N θ ( m − , t m − + v H θ ( m − , t m − )= N θ ( m, t m + v H θ ( m − , t m − . The proof of the following lemma is very similar.
Lemma 4.12.
For all m ≥ , we have H s θ ( m, = N s θ ( m, + v H s θ ′ ( m − , . Lemma 4.13.
For all m ≥ the element H s s s θ ( m, is equal to N s s s θ ( m, + v ⌊ m − ⌋ X i =0 v i ( N θ ( m − i, + N s θ ′ ( m − − i, ) + ⌊ m ⌋ X i =1 v i ( v − N θ ( m − i, + N s θ ( m − i, ) . (4.29)22 roof. For m = 0 and m = 1 the result can be checked by a direct computation. We then assume m ≥ .Definition 4.1 and Theorem 4.5 give us the following identity H θ ( m, = ⌊ m ⌋ X i =0 v i N θ ( m − i, . Then, combining (4.14), (4.18) and (4.22) we get H s s s θ ( m, = (cid:0) H s H s H s − H s − H s (cid:1) H θ ( m, = (cid:0) H s H s H s − H s (cid:1) H θ ( m, − ( v + v − ) H θ ( m, = (cid:0) H s H s H s − H s (cid:1) ⌊ m ⌋ X i =0 v i N θ ( m − i, − ( v + v − ) ⌊ m ⌋ X i =0 v i N θ ( m − i, = ⌊ m ⌋ X i =0 (cid:0) H s H s H s − H s (cid:1) v i N θ ( m − i, − ( v + v − ) ⌊ m ⌋ X i =0 v i N θ ( m − i, . We assume m is odd, the case m even being similar. We then use Lemma 3.19 to conclude that H s s s θ ( m, = ⌊ m ⌋ X i =0 v i (cid:2) N s s s θ ( m − i, − v N s s s θ ( m − i +1) , (cid:3) + ⌊ m ⌋ X i =0 v i (cid:2) v ( N θ ( m − i, + N s θ ′ ( m − − i, ) + v ( v − N θ ( m − i +1) , + N s θ ( m − i +1) , ) (cid:3) . (4.30)We notice that the first sum in (4.30) telescopes to N s s s θ ( m, and therefore the right-hand side of (4.30)reduces to (4.29). We stress that the apparent discrepancy between both expressions is solved by the factthat in (4.30) the term v − N θ ( m − i +1) , + N s θ ( m − i +1) , becomes zero when i = ( m − / . Corollary 4.14.
For all m ≥ we have H s s s θ ( m, t m = N s s s θ ( m, t m + v N s s s θ ( m − , t m − + v k +2 N s s + m X i =0 v i +1 N θ ( m − i, t m − i + m X i =1 v i N s θ ′ ( m − i, t ′ m − i + m X i =2 (cid:0) v i − N θ ( m − i, t m − i + v i N s θ ( m − i, t m − i (cid:1) . (4.31) Consequently, H s s s θ ( m +1 , t m +1 = N s s s θ ( m +1 , t m +1 + v H s s s θ ( m, t m − v N s s s θ ( m − , t m − + v N θ ( m +1 , t m +1 + v N s θ ′ ( m, t ′ m + v N θ ( m − , t m − + v N s θ ( m − , t m − . (4.32) Proof.
We first notice that ( s s s θ ( m, − belongs to C (resp. ϕ ( C ) ) if m is even (resp. odd). Thus, usingeither (4.17) or its image under ϕ we conclude that H t m H ( s s s θ ( m, − = H t m ( s s s θ ( m, − . It follows that H s s s θ ( m, t m = H s s s θ ( m, H t m . Therefore we can use the expression for H s s s θ ( m, given in Lemma 4.13 in order to compute H s s s θ ( m, t m . Then (4.31) follows by a combination of Lemma3.7 with n = 1 , Lemma 3.9, Lemma 3.11 and its ϕ -image and Lemma 3.12. Finally, (4.32) is a directconsequence of (4.31).The following result can be proved with similar arguments. For the sake of brevity we omit the proof. Lemma 4.15.
Let m ∈ N . Then, we have H s θ ( m +1 , t m +1 = N s θ ( m +1 , t m +1 + v N s θ ′ ( m, t ′ m + v H s θ ( m, t m , H s θ ′ ( m +1 , t ′ m +1 = N s θ ′ ( m +1 , t ′ m +1 + v N s θ ( m, t m + v H s θ ′ ( m, t ′ m . Kazhdan-Lusztig polynomials for the thick region
In this section we compute Kazhdan-Lusztig basis elements corresponding to elements in the thick region. Werecall from Figure 1 that this region is made of four sub-regions: north ( N ), south ( S ), east ( E ) and west ( W ).Recall from introduction that N = { x n | n ≥ } ∪ { x n | n ≥ } , where x n = a · · · a n and x n = s s s x n − ,and the a i are defined by the sequence { a n } ∞ n =1 = ( s , s , s , s , s , s , s , s , s , s , s , s , . . . ) . We also recallthat S = ϕ ( N ) , E = { e n = s x ′ n | n ≥ } ∪ { e ′ k | k ≥ } , and W = s ( E ) ∪ { s } . Definition 5.1.
We recall from the introduction the notation f ( k ) = 3 k + 1 and u n = s x n . For all k ≥ we define ˆ H x f ( k ) = N x f ( k ) + v N x f ( k − + k − X j =2 v j − ( N e f ( k − j ) + N u f ( k − j ) ) + v k − N s s . (5.1) Lemma 5.2.
Let Y = H s H s H s − H s . For all k ≥ we have ˆ H x f ( k ) Y = ˆ H x f ( k +1) + ˆ H x f ( k − + H s θ ′ ( k − , t ′ k − + H θ ( k − , t k − + H θ ( k − , t k − + H s s s θ ( k − , t k − . Proof.
We proceed by induction on k . The case k = 2 follows by a direct computation. We assume thelemma holds for some k ≥ . It follows directly from Definition 5.1 that ˆ H x f ( k +1) = N x f ( k +1) + v N e f ( k − + v N u f ( k − − v N x f ( k − + v ˆ H x f ( k ) . (5.2)Multiplying this equality on the right by Y and using our inductive hypothesis and Lemma 3.20 we obtainthat ˆ H x f ( k +1) Y is equal to the sum of all the entries of the following matrix. N x f ( k +2) N x f ( k ) N θ ( k, t k N s s s θ ( k − , t k − N s θ ′ ( k − , t ′ k − v N θ ( k − , t k − v N u f ( k ) v N u f ( k − v N θ ( k − , t k − v N θ ( k − , t k − N θ ( k − , t k − v N u f ( k − v N e f ( k ) v N e f ( k − v N s θ ( k − , t k − v N s θ ′ ( k − , t ′ k − v N s θ ( k − , t k − v N s θ ′ ( k − , t ′ k − − v N x f ( k ) − v N x f ( k − − v N θ ( k − , t k − − v N s s s θ ( k − , t k − − v N s θ ′ ( k − , t ′ k − − v N θ ( k − , t k − v ˆ H x f ( k +1) v ˆ H x f ( k − v H s θ ′ ( k − , t ′ k − v H θ ( k − , t k − v H θ ( k − , t k − v H s s s θ ( k − , t k − We denote this matrix by A = ( A ij ) . Using (5.2) we see that ˆ H x f ( k +2) = X i =1 A i and ˆ H x f ( k ) = X i =1 A i . By Lemma 4.11 we have H θ ( k, t k = A + A and H θ ( k − , t k − = A + A . On the other hand, Lemma 4.15implies that H s θ ′ ( k − , t ′ k − = A + A + A . We also observe that A + A = A + A = A + A = 0 ,this last equality being a consequence of the fact that θ ( k − , t k − = u f ( k − . Finally, we notice thatCorollary 4.14 yields H s s s θ ( k − , t k − = A + A + A + A + A + A + A . Putting all this togetherwe obtain ˆ H x f ( k +1) Y = ˆ H x f ( k +2) + ˆ H x f ( k ) + H s θ ′ ( k − , t ′ k − + H θ ( k, t k + H θ ( k − , t k − + H s s s θ ( k − , t k − , as we wanted to show. Theorem 5.3.
For all k ≥ we have H x f ( k ) = ˆ H x f ( k ) .Proof. For k = 2 the result follows by a direct computation. We now fix some k ≥ and assume the theoremholds for all ≤ k ′ ≤ k . We notice that G x f ( k +1) ( ˆ H x f ( k +1) ) = 1 and G z ( ˆ H x f ( k +1) ) ∈ v N [ v ] , for all z < x f ( k +1) .Thus we only need to show that ˆ H x f ( k +1) is self-dual. This is an immediate consequence of Lemma 5.2 andour inductive hypothesis. Theorem 5.4.
Let k ≥ . We have H x k +1 H s = H x k +2 . (5.3)24 x k +2 H s = ( H x k +3 + H x k +1 + H θ ( k − , + H s θ ′ ( k − , + H θ ( k − , , if k is even; H x k +3 + H x k +1 + H s s s θ ( k − , , if k is odd. (5.4) H x k +2 H s = ( H x k +3 + H x k +1 + H s s s θ ( k − , , if k is even; H x k +3 + H x k +1 + H θ ( k − , + H s θ ′ ( k − , + H θ ( k − , , if k is odd. (5.5) Proof.
The claim can be checked directly for k = 2 . From now on we assume k ≥ . An inspection of theexplicit formula for H x k +1 provided in Definition 5.1 allows us to conclude that xs > x for all x ∈ W suchthat µ ( x, x k +1 ) = 0 . Thus the sum in (3.1) is empty and (5.3) follows.We now prove (5.4). We first treat the case k even. Equation (3.2), Lemma 5.2 and Theorem 5.3 imply H x k +1 H s H s H s = ( v + v − ) H x k +1 + H x k +4 + H x k − + H s θ ′ ( k − , s + H θ ( k − , s + H θ ( k − , s + H s s s θ ( k − , s . (5.6)On the other hand, we can combine (3.4), (5.1) and (5.3) to obtain h s θ ′ ( k − , ,x k +2 ( v ) = v − h s θ ′ ( k − , ,x k +1 ( v ) + h e k − ,x k +1 ( v )= v − ( v ) + v = v + v h θ ( k − , ,x k +2 ( v ) = v − h θ ( k − , ,x k +1 ( v ) + h x k − ,x k +1 ( v )= v − ( v + v + v ) + ( v + v + 2 v )= v + 3 v + 2 v + v . From this we conclude that µ ( s θ ′ ( k − , , x k +2 ) = µ ( θ ( k − , , x k +2 ) = 1 . (5.7)We have the following equalities H x k +1 H s H s H s = H x k +2 H s H s = H x k +3 + H x k +1 + H θ ( k − , + H s θ ′ ( k − , + H θ ( k − , + X y ∈ Y m y H y H s = H x k +4 + H s s s θ ( k − , s + X z ∈ Z n z H z ! + ( v + v − ) H x k +1 + H θ ( k − , s + H s θ ′ ( k − , s + H θ ( k − , s + X y ∈ Y m y H ys . (5.8)where m y , n z ∈ N , Z = { z ∈ W | D R ( y ) = { s , s }} and Y = { y ∈ W | D R ( y ) = { s , s }} \ { s θ ′ ( k − , , θ ( k − , } . Let us explain how to obtain the above equalities. The first equality is a direct consequence of (5.3). Forthe second one we use (3.1); first consider the elements in ⋖ x k +2 satisfying ws < w . Lemma 2.13 togetherwith an easy case analysis show that these elements are x k +1 and θ ( k − , . This justifies the occurrence ofthe terms H x k +1 and H θ ( k − , . On the other hand, (5.7) explains the occurrence of the terms H s θ ′ ( k − , and H θ ( k − , . Finally, the equality D R ( x k +2 ) = { s } allows us to conclude, via Lemma 3.6, that any otherterm occurring must have right descent set equal to { s , s } . This explains the appearance of the sum. Weremark that D R ( s θ ′ ( k − , D R ( θ ( k − , { s , s } as well (for k even). However the elements s θ ′ ( k − , and θ ( k − , were already considered, which explains the somewhat strange definition of Y .For the last equality, the terms inside the parentheses are justified applying the same arguments used forthe second equality to the multiplication H x k +3 H s . We notice that D R ( x k +3 ) = { s } and this explainsvia Lemma 3.6 the definition of Z . The term ( v + v − ) H x k +1 follows by (3.2). Finally, the occurrence of all25he other terms follows by applying Remark 4.9, which (in its ϕ -version) is equivalent to the statement thatif D L ( w ) = { s , s } then H s H w = H s w . Using inverses to move from left to right, the latter is equivalentto saying that if D R ( w ) = { s , s } then H w H s = H ws . This is the version we need to conclude.A comparison of the expressions for H x k +1 H s H s H s given in (5.6) and (5.8) yields H x k − = X z ∈ Z n z H z + X y ∈ Y m y H ys . Suppose that x k − = ys for some y ∈ Y . Multiplying by s on the right we obtain that x k − ∈ Y since k is even. However, D R ( x k − ) = { s } and we reach a contradiction. We conclude that m y = 0 for all y ∈ Y .Therefore, H x k +2 H s = H x k +3 + H x k +1 + H θ ( k − , + H s θ ′ ( k − , + H θ ( k − , , as we wanted to show. This completes the proof of (5.4) for k even.We now assume k is odd. As before, Equation (3.2), Lemma 5.2 and Theorem 5.3 imply H x k +1 H s H s H s = ( v + v − ) H x k +1 + H x k +4 + H x k − + H s θ ′ ( k − , s + H θ ( k − , s + H θ ( k − , s + H s s s θ ( k − , s . (5.9)On the other hand, arguing as in (5.8) we obtain H x k +1 H s H s H s = H x k +2 H s H s = H x k +3 + H x k +1 + H s s s θ ( k − , + X y ∈ Y m y H y H s = H x k +4 + H θ ( k − , s + X z ∈ Z n z H z + ( v + v − ) H x k +1 + H s s s θ ( k − , s + X y ∈ Y m y H ys . (5.10)where m y , n z ∈ N , Z = { z ∈ W | D R ( y ) = { s , s }} and Y = { y ∈ W | D R ( y ) = { s , s }} .A comparison of the expressions for H x k +1 H s H s H s given in (5.9) and (5.10) yields H x k − + H s θ ′ ( k − , s + H θ ( k − , s = X z ∈ Z n z H z + X y ∈ Y m y H ys . As before we want to show that the sum over Y is zero. Suppose that x k − = ys for some y ∈ Y .Multiplying by s on the right we have that x k − ∈ Y since k is odd. Since D R ( x k − ) = { s } we obtaina contradiction. We conclude that term H x k − does not come from the sum over Y . We now suppose that s θ ′ ( k − , s = ys for some y ∈ Y . Multiplying by s on the right we obtain y = s θ ′ ( k − , s s = s θ ′ ( k − , s s s s s = s θ ′ ( k − , s s s . It follows that s θ ′ ( k − , s s s ∈ Y . However, D R ( s θ ′ ( k − , s s s ) = { s } and we reach a contra-diction. We conclude that term H s θ ′ ( k − , s does not come from the sum over Y . Similarly, we can provethat H θ ( k − , s does not come from the sum over Y . It follows that the sum over Y is zero. Therefore, H x k +2 H s = H x k +3 + H x k +1 + H s s s θ ( k − , , as we wanted to show.The proof of (5.5) is dealt with similarity. The details are left to the reader.Theorem 5.3 and Theorem 5.4 give formulas for all Kazhdan-Lusztig basis elements indexed by elementsof N of length greater than six; the remainder can be obtained by direct calculation. As we already pointedout at the beginning of this section, S = ϕ ( N ) , therefore acting by ϕ in each one of these formulas we obtainthe corresponding formulas for all the Kazhdan-Lusztig basis elements indexed by elements of S .26 .2 East and West. Definition 5.5.
For k ≥ we define the following elements ˆ H e k +1 = k X i =0 v i N e k − i )+1 , (5.11) ˆ H e k +2 = N e k +2 + k X i =1 v i N e k − i )+2 , (5.12) ˆ H e k +3 = N e k +3 + v N e ′ k + k X i =1 v i N e k − i )+1 + k X i =1 v i N ϕ i ( s θ ( k − i − , . Lemma 5.6.
Let k ≥ . We have ˆ H e f ( k ) H s H t k H t ′ k = ˆ H e f ( k +1) + ( v + v − ) ˆ H e f ( k ) + ˆ H e f ( k − + H s θ ( k − , t ′ k + H s θ ′ ( k − , t k . (5.13) Proof.
We proceed by induction on k . A direct computation gives us the result for k = 1 . We fix k ≥ andassume that (5.13) holds for all k ′ ≤ k . By Definition 5.5 we have ˆ H e f ( k +1) = N e f ( k +1) + v ˆ H e f ( k ) . (5.14)Then, multiplying this last equation on the right by H s H t k H t ′ k and using Lemma 3.20, Lemma 3.4 and ourinductive hypothesis we conclude that ˆ H e f ( k +1) H s H t k H t ′ k is equal to the sum of the entries of the followingmatrix N e f ( k +2) ( v + v − ) N e f ( k +1) N e f ( k ) N s θ ( k, t k N s θ ′ ( k, t ′ k v ˆ H e f ( k +1) v ( v + v − ) ˆ H e f ( k ) v ˆ H e f ( k − v N s θ ′ ( k − , t ′ k − v N s θ ( k − , t k − v H s θ ( k − , t ′ k v H s θ ′ ( k − , t k Bearing in mind that t ′ k = t k − , Lemma 4.15 together with (5.14) show that if we add the entries of theabove matrix by columns then we obtain ˆ H e f ( k +1) H s H t k H t ′ k = ˆ H e f ( k +2) + ( v + v − ) ˆ H e f ( k − + ˆ H e f ( k ) + H s θ ( k, t ′ k +1 + H s θ ′ ( k, t k +1 . This completes our induction.
Theorem 5.7.
For all k ≥ and j ∈ { , , } we have ˆ H e k + j = H e k + j . (5.15) Proof.
We split the proof of (5.15) in three cases in accordance with the value of j . Case A. ( j = 1) For k = 1 the result can be checked by hand. We assume k > . It follows directlyfrom the definition of ˆ H e k +1 that G e k +1 ( ˆ H e k +1 ) = 1 and G z ( ˆ H e k +1 ) ∈ v N [ v ] , for all z < e k +1 . Therefore,we only need to check that ˆ H e k +1 is self-dual. The latter follows by an inductive argument using Lemma 5.6. Case B. ( j = 2) We will prove this identity by showing that c ( ˆ H e k +2 ) = c ( H e k +2 ) and that H e k +2 H ≥ ˆ H e k +2 . Inspection of the formula (5.11) and the set ⋖ e k +1 from Lemma 2.14 allows us to conclude theidentity H e k +2 = H e k +1 H s , applying (3.1) and Case A . Therefore, the first step reduces to showing that c ( ˆ H e k +2 ) = 2 c ( H e k +1 ) which follows from a straightforward computation using Lemma 2.12.In order to prove that H e k +2 H ≥ ˆ H e k +2 , by degree reasons and the monotonicity of H e k +2 , it is enoughto check that G e k − i )+2 ( H e k +2 ) = G e k − i )+2 ( H e k +1 H s ) ≥ v i for ≤ i ≤ k . In view of (5.11), we see that G e k − i )+2 ( H e k +1 H s ) ≥ G e k − i )+2 (( v i H e k − i )+1 + v i +1 H e k − i )+2 ) H s ) = 2 v i as desired. 27 ase C. ( j = 3) Note that e k +3 = e k +2 t k . Proceeding as in Case B , inspection of formula (5.12)and Lemma 2.14 allows us to conclude H e k +3 = H e k +2 H t k − H e k +1 − H s θ ( k − , . (5.16)Combining the facts that c ( H e k +2 H t k ) = c ( H e k +1 H s H t k ) = 4 c ( H e k +1 ) and c ( H s θ ( k − , ) = 2 c ( H θ ( k − , ) ,a straightforward calculation using Lemma 2.12, Lemma 2.7, Theorem 4.5 and Corollary 2.5 shows that c ( H e k +3 ) = c ( ˆ H e k +3 ) .Next we show that H e k +3 H ≥ ˆ H e k +3 . By degree reasons and monotonicity of H e k +3 , it suffices to provethat G e ′ k ( H e k +3 ) ≥ v , G e ( H e k +3 ) ≥ v k , and for ≤ i ≤ k − , H e k +3 H ≥ v i N e k − i )+1 + v i N ϕ i ( s θ ( k − − i, . (5.17)We will only show (5.17), as the other items follow by similar (and easier) arguments.Observing that ( ≤ e k − i )+1 ) ∩ ( ≤ ϕ i ( s θ ( k − − i, ≤ ϕ i ( e ′ k − i ) ) , the proof of (5.17) reduces to showing that G e k − i )+1 ( H e k +3 ) ≥ v i ,G ϕ i ( s θ ( k − − i, ( H e k +3 ) ≥ v i , and G ϕ i ( e ′ k − i ) ) ( H e k +3 ) ≥ v i +1 . The term v i N e k − i )+2 in (5.12) has three summands of interest: v i +1 H e k − i )+1 , v i +1 H ϕ i ( s θ ( k − − i, ,and v i +2 H ϕ i ( e ′ k − i ) ) . Multiplying these terms by H t k contributes v i H e k − i )+1 , v i H ϕ i ( s θ ( k − − i, , and v i +1 H ϕ i ( e ′ k − i ) ) to H e k +3 . Then using (5.11) and Lemma 4.12 with (5.16), the subtraction of terms v i N e k − i )+1 and v i N ϕ i ( s θ ( k − − i, has the only effect in the degrees of interest, leaving us with the desiredcoefficients.Theorem 5.7 provides formulas for all the Kazhdan-Lusztig basis elements indexed by elements locatedin E of length at least five. Smaller elements can be computed directly. The following theorem providesformulas for all the Kazhdan-Lusztig basis elements indexed by elements located in W , thus completing ourdescription of Kazhdan-Lusztig basis for the thick region. Theorem 5.8.
Let n ≥ . Then H w n = H s H e n .Proof. We notice that D L ( e n ) = { s , s } . Then, Lemma 3.6 implies H s H e n = H w n + X z ⋖ e n s z
N E , SW and SE ) the sub-region of the thin regionformed by triangles located to the northwest (resp. northeast, southwest and southeast) of the identitytriangle. We define d n as the product of the first n symbols of the infinite string s s s s s s s s s s s s . . . .The elements d n comprise the thin wall that goes towards the northwest in our convention, and the elements d ′ n = ϕ ( N W ) are the southwestward thin wall (see Figure 1). In other words, N W = { d n | n ≥ } and SW = { d ′ n | n ≥ } . Finally, we define d n = s d n and with these describe the remaining two thin walls as SE = { d n | n ≥ } and N E = { d ′ n | n ≥ } .The thin region coincides very nearly with the two-sided cell of W consisting of the elements with uniquereduced expression. Wang has shown [Wan11] that there exist u ∈ W for which µ ( u, w ) = 0 for infinitely28any w in the thin region, meaning that the W -graph of the group is not locally finite. According to ourconjecture, there are in fact infinitely many such u . This indicates an additional level of complexity for theKazhdan-Lusztig basis in the thin region, reflected in the formulas below.We first define some notation: For x, z ∈ W , let D zx := X w ≤ xw z v l ( x ) − l ( w ) H w Notice that D zx is a truncation of N x . We further define U x := N x + D xx ′ = U x ′ . Conjecture 6.1.
For all k ≥ ,1. H d k +3 = N d k +3 + v D s s θ ′ (0 ,k − s s θ (0 ,k − + ( v + v ) H θ (0 ,k − + v H s e + v k X i =3 (cid:16) H θ (0 ,k − i ) + H θ ′ (0 ,k − i ) (cid:17) + v k X i =2 (cid:16) U s s θ ′ (0 ,k − i ) s s + U θ (0 ,k − i ) s s + H s s θ (0 ,k − i ) + H s s θ ′ (0 ,k − i ) (cid:17) , H d k H s = H d k +1 + H d k − + k − X i =0 H θ (1 ,i ) + k − X i =0 H s s θ (1 ,i ) . H d k +1 H s = H d k +2 + k − X i =0 H θ (0 ,i ) + k − X i =0 H s s θ (0 ,i ) . H d k +2 H s = H d k +3 + H d k +1 + k − X i =0 H s s θ ′ (1 ,i ) + k − X i =0 H θ ′ (1 ,i ) H d k +3 H s = H d k +4 + k − X i =0 H s s θ ′ (0 ,i ) + k − X i =0 H θ ′ (0 ,i ) . H s H d k = H d k + k − X i =0 (cid:16) H θ ′ (0 ,i ) s s s + H θ ′ (0 ,i ) s (cid:17) H s H d k +1 = H d k +1 + k − X i =0 H θ ′ (0 ,i ) + k − X i =0 H θ ′ (0 ,i ) s s H s H d k +2 = H d k +2 + H θ ′ (0 ,k − s + k − X i =0 (cid:16) H θ ′ (0 ,i ) s + H θ ′ (1 ,i ) (cid:17) H s H d k +3 = H d k +3 + k − X i =0 H θ ′ (0 ,i ) s s + k − X i =0 H θ ′ (0 ,i ) These formulas above have been checked up to k = 5 . Note again that these formulas cover both thenorthwest and southeast walls of the thin region, while basis elements from the other walls of this region canbe obtained by applying the automorphism ϕ . References [BB06] Anders Bjorner and Francesco Brenti.
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