ROUQUIER COMPLEXES DONGKWAN KIM

ROUQUIER COMPLEXES DONGKWAN KIM 1. Rouquier complexes 1.1. Setup. In this talk k is a field of characteristic 0. Let (W, S) be a Coxeter group with S = n <. Also we let V = s S ke s be the reflection representation of W, and R = k[v ]. R is a graded algebra with V in degree 2. We abbreviate M R N = MN for a right R-module M and a left R-module N. We let {α s } s S be the dual basis of {e s } s S, which can be considered as elements in V R. Thus R k[α 1,, α n ] is a polynomial algebra of n variables. The natural action of W on V induces that on R, i.e. (w f)(v) := f(w 1 (v)), and thus on R mod gr (the category of graded R-module) and D b (R mod gr ) (its bounded derived category) in a way that for f R and m w M we have f m = (w 1 f)(m). Note that for M R mod gr, w M R w M, where R w is a R-bimodule such that R w R as a left R-module and m a = mw(a) for m R w, a R, i.e. the right action is twisted by w. In other words, W End(D b (R mod gr )) : w R w R is a well-defined action of W on D b (R mod gr ). Also note that for any w, w W, we have a canonical isomorphism R w R w R ww : 1 1 1 of R-bimodules. Let K b (R mod gr ) be the bounded homotopy category of R mod gr. This is the category of bounded chain complexes in R mod gr modulo homotopy equivalence: for f, g : A B, we say that f is homotopic to g, denoted f g, if there exists h : A B 1 such that f g = dh + hd. In K b (R mod gr ), A B if there exist f : A B and g : B A such that gf id A and fg id B. Note that this category is well-defined if one replaces R mod gr by any additive category, not necessarily abelian. In particular, if we let SBim gr be the category of graded Soergel bimodules and R bim gr be the category of graded R-bimodules, then K b (SBim gr ) and K b (R bim gr ) are well-defined and we havek b (SBim gr ) K b (R bim gr ). 1.2. Definition of Rouquier complexes. Our goal is to lift the action of W on D b (R mod gr ) mentioned above to K b (R mod gr ), using Soergel bimodules. How can we lift R w? We may consider: W End(K b (R mod gr )) : w R w R which gives a trivial lift, but this is not interesting... Here we introduce another interesting lifting introduced by Rouquier. However, instead we should work with the braid group B W of (W, S) instead of W itself. Date: November 5, 2017. 1

ROUQUIER COMPLEXES 2 Any object M K b (R bim gr ) defines an endofunctor K b (R mod gr ) K b (R mod gr ) : N M R N =: MN. For a word w with alphabet S S 1, we will define an invertible object F w K b (SBim gr ) K b (R bim gr ) which descends to R w D b (R bim gr ) (up to a shift). Here, w W is w considered as a product of simple reflections in W. First we recall some morphisms defined in Boris talk: m s : B s R(1) : f g fg, j s : B s B s (1) B s : f g h s (g)f h m a s : R( 1) B s : f fα s 1 + f α s j a s : B s (1) R : f g f 1 g ((i) is the grading shift as a graded R-bimodule.) For s S, we define F s := 0 B s m s R(1) 0 K b (SBim gr ) F s 1 := 0 R( 1) ma s B s 0 K b (SBim gr ) The box indicates cohomology degree 0. From the following exact sequences of graded R-bimodules it easily follows that However, they are not equal in K b (R bim gr ). 0 R s ( 1) f f αs fαs 1 m B s s R(1) 0 0 R( 1) ma s f g fs(g) B s R s (1) 0 R s F s (1) F s 1( 1) in D b (R bim gr ). Exercise 1.1. Prove that R s F s (1) and R s F s 1( 1). In general, for a word w = abcd with alphabet S S 1, we define F w := F a F b F c F d K b (SBim gr ), called the Rouquier complex corresponding to w. In particular, we have F = R, F s = F s, and F s 1 = F s 1. Remark. Indeed, the split Grothendieck group of SBim gr is isomorphic to the Hecke algebra of W. Under this correspondence, the image of the class of F s is the standard basis T s. 1.3. Some properties of Rouquier complexes. We claim that indeed this correspondence w F w defines a weak action of B W on K b (R mod gr ). For that, we need to check that F w1 F w2 if w 1 and w 2 represent the same element in B W. (Note that this does not hold if we only assume that w 1 and w 2 represent the same element in W, e.g. F s F s 1.) To that end, it suffices to show the following. (1) F s F s 1 F s 1F s R for s S. (2) F s F t F s F t F s F t where each side is the product of m terms, for s, t S and m N such that (st) m = 1 W.

ROUQUIER COMPLEXES 3 First we show that F s F s 1 R in K b (R bim gr ). The tensor product of F s and F s 1 is described as follows. (Here, we understand this diagram as a single chain complex, each of whose terms is the direct sum of objects on each column.) 1 m a B s B s F s F s 1 = B s ( 1) B s (1) m R m a m s 1 We claim that it is homotopy equivalent to R. Indeed, if we define (we omit the subscript s) (1) ι 1 = (1 m a, m) : B s ( 1) B s B s R (2) ι 2 = (j a, 0) : B s (1) B s B s R (3) ι 3 = (j a m a, Id) : R B s B s R (4) ρ 1 = j 0 : B s B s R B s ( 1) (5) ρ 2 = (m 1) ( m a ) : B s B s R B s (1) (6) ρ 3 = mj Id : B s B s R R then they give a decomposition of B s B s R into B s ( 1) B s (1) R, i.e. ρ i ι i = Id and ρ j ι i = 0 if i j. (1 m a,m) (m 1) ( m a ) B s ( 1) B s B s R B s (1) j 0 (j a,0) (j a m a,id) R mj id In other words, the following diagram describes that F s F s 1 R in K b (R bim gr ). (One can easily show that the vertical maps are mutual inverses using ι i and ρ i for 1 i 3.) F s 1F s R can be proved similarly. (1 m a,m) B s ( 1) B s B s R B s (1) mj id (m 1) ( m a ) 0 R 0 (j a m a,id) (1 m a,m) (m 1) ( m a ) B s ( 1) B s B s R B s (1)

ROUQUIER COMPLEXES 4 However, F s F s is not homotopy equivalent to R, even up to a shift. Indeed, we have 1 m B s (1) F s F s = B s B s R(2) m 1 B s (1) and it is homotopy equivalent to the following chain complex. m 1 1 α B s ( 1) s 1 1 α s m B s (1) R(2) As B s B s B s (1) B s ( 1), the claim follows if we can cancel out B s (1) in cohomological degree 0 and 1, respectively. To be more precise, we use the following diagram. (The vertical maps are mutual inverses in K b (R bim gr ).) (1 m,m 1) B s B s B s (1) B s (1) R(2) m m ( m) j a (f,g) f g Id 1 1 α s 1 1 α s B s ( 1) B s (1) R(2) m 1 1 α s 1 1 1 α s 1 id 0 Id (1 m,m 1) Also, there exists a quasi-isomorphism m ( m) B s B s B s (1) B s (1) R(2) 1 1 α s 1 1 α s B s ( 1) B s (1) R(2) m m a R( 2) Thus, F s F s R( 2) in D b (R bim gr ). However, it is not a homotopy equivalence. Indeed, otherwise we would have F s F s 1( 2), but it can be easily seen to be false. Exercise 1.2. Prove that F s F s 1( 2). (Hint: try to construct homotopy equivalences between F s and F s 1( 2). What are the degree zero endomorphisms of B s?) Exercise 1.3. Show that F s F s F s (m terms) is homotopy equivalent to d B s (1 m) 0 d B s (3 m) 1 d m 2 B s (m 1) m R(m) where differentials are given by d m 2, d m 4, = 1 1 α s 1 1 α s, d m 3, d m 5, = 1 1 α s 1 + 1 α s. Now we assume s, t S are given. If m st = 2, then indeed it is not hard to show that F s F t F t F s K b (R bim gr ), since B s B t B t B s B st. What if m st = 3? We want to show that F s F t F s F t F s F t K b (R bim gr ). Equivalently, we will show that F s [1]F t [1]F s [1]

ROUQUIER COMPLEXES 5 F t [1]F s [1]F t [1], where [i] is the cohomology shift. 1 cube-shaped diagram. F s [1]F t [1]F s [1] is expressed as the following B s B t (1) B s (2) B s B t B s B s B s (1) B t (2) R(3) B t B s (1) B s (2) The solid arrows are multiplication maps and the dashed ones are their negatives. From [EK10], we have a homotopy equivalence which is denoted by squiggling arrows in the following diagram. B s B t (1) B s (2) B s B t B s B s B s (1) B t (2) R(3) B t B s (1) B s (2) s fs g s Id B t B s (1) B t (2) B t B s B t B t B t (1) B s (2) R(3) Here, s, f s, g s are defined as follows. B s B t (1) B t (2) s is the projection B s B t B s B sts composed with the injection B sts B t B s B t. It satisfies 1 1 1 1 1 1 1 1, 1 (2α s + α t ) 1 1 (α s + 2α t ) 1 1 1 + 1 1 1 (α s + 2α t ). Note that 1 1 1 1, 1 (2α s + α t ) 1 1 generate B s B t B s as an R-bimodule. The reason we choose 1 (2α s +α t ) 1 1 as a second generator is that 1 (2α s +α t ) 1 1 = 1 1 (2α s +α t ) 1, i.e. 2α s + α t is fixed by t. Also, we have that t s = (1 e s ) = Id + m a t j a s j s m t, which is the projection idempotent B s B t B s B sts defined in Boris talk. (There, we defined e = m a t j a s j s m t and the corresponding idempotent was written 1 e.) 1 This shift is introduced so that we can directly use the result of [EK10]. If one wants to compute FsF tf s, signs of some maps should be changed accordingly.

0 m a t j s Id f s = (1 x)m s jt a jt a m a t m s j s xjt a m s Id j s m a t 0 0 (1 x)id 0 g s = Id 0 Id 0 xid 0 ROUQUIER COMPLEXES 6 Here, x k is arbitrary. The homotopy inverse is similarly defined, by switching s and t. We need to show that their composition is homotopy equivalent to the identity; the corresponding chain homotopy is given by the arrows with double lines in the following diagram. B s B t (1) B s (2) B s B t B s B s B s (1) B t (2) R(3) B t B s (1) B s (2) h 3 t s=1 e f tf s h 2 g tg s Id h 1 B s B t (1) B s (2) B s B t B s B s B s (1) B t (2) R(3) B t B s (1) B s (2) Here, we have 0 m a s j t Id 0 m a f t f s = (1 x)m t js a js a m a sm t j t xjs a t j s Id m t (1 x)m s j Id j t m a t a jt a m a t m s j s xjt a m s s 0 Id j s m a t 0 Id j s m a t 0 = xj s a m t 2(1 x)l αt js a j s + 2xjs a j s R αt (1 x)m t js a, 0 m a t j s Id 0 (1 x)id 0 0 (1 x)id 0 (1 x)id 0 (1 x)id g t g s = Id 0 Id Id 0 Id = 0 Id 0. 0 xid 0 0 xid 0 xid 0 xid Here, L αt, R αt are left and right multiplication by α t, respectively. Chain homotopy maps are defined as follows. (y k is arbitrary.) h 1 = [ 0 m a t js a j s 0 ], h 2 = 0 0 0 yj s 0 (1 y)j s, h 3 = 0 0 0 0

ROUQUIER COMPLEXES 7 This proves that F s F t F s F t F s F t for m st = 3. See [Rou04, Section 3] for the proof in general cases. Remark. Indeed, F s F t F s is homotopy equivalent to B s B t (1) B s (2) B sts R(3) B t B s (1) B t (2) with certain chain maps, which are symmetric with respect to s and t. Since B sts B tst, the relation F s F t F s F t F s F t follows directly. However, it is not easy to show that the chain above is indeed homotopy equivalent to F s F t F s. In sum, we have the following theorem. Theorem 1.4. [Rou04, Proposition 3.4] The map s F s extends to a morphism from B W to the group of isomorphism classes of invertible objects of K b (R bim gr ). Furthermore, its image lies in K b (SBim gr ), the category of graded Soergel bimodules. 1.4. A strict monoidal category B W. In fact, we can proceed further; for w B W, let t 1 t r, u 1 u s be two words of w with alphabet S S 1. Then F t1 F tm F u1 F un, and furthermore Thus the canonical map Hom Kb (R bim gr)(f t1 F tm, F u1 F un ) End K b (R bim gr)(r) k Hom Db (R bim gr)(f t1 F tm, F u1 F un ) End Db (R bim gr)(r) k Hom Kb (R bim gr)(f t1 F tm, F u1 F un ) Hom Db (R bim gr)(f t1 F tm, F u1 F un ) is an isomorphism. Recall that we already have a canonical isomorphism R t1 R tm R u1 R un : 1 1 1 1. For each w B W, we consider the system {F t1 F tm } of functors with isomorphisms F t1 F tm F u1 F un which descends to such a canonical isomorphism as above. We define G w as the limit of this direct system, which is unique up to a unique isomorphism. Then for w, w B W, there are unique isomorphisms G w G w G w w and G R. We consider the full subcategory B W of K b (R bim gr ) with objects {G w } w BW. Also define G v := G v 1. Then, we have the following theorem. Theorem 1.5. [Rou04, Theorem 3.7] The category B W is a strict rigid monoidal category, and w G w gives a strong action of B W on K b (R mod gr ). As a result, the decategorification of B W is a quotient of B W. Rouquier conjectured that it is indeed the same as B W, i.e. the action is faithful. Some partial answers to this conjecture are known: [KS02] for type A, [BT11] for type ADE, and [Jen17] for finite type in general.

ROUQUIER COMPLEXES 8 2. Khovanov-Rozansky homology and knot invariants In this section, we relate Rouquier complexes and Khovanov-Rozansky homology, which is a certain invariant of links. We will see that one can calculate HOMFLY-PT polynomials from such complexes. 2.1. Hochschild homology. We define M R := M/[R, M] = R R en M, i.e. taking coinvariants. Here, R en := R k R is the enveloping algebra of R. Then the functor M M R is right exact and we define HH i (R, M) := Tor Ren i (R, M), HH(R, M) := i 0 HH i (R, M). called Hochschild homology of M. Similarly, one can define Hochschild cohomology as a derived functor of taking invariants. Indeed, for a polynomial algebra R we have HH i (R, M) HH n i (R, M), which comes from the self-duality of the Koszul resolution of R. But this isomorphism does not hold for general R. How can we calculate such homology? Recall that R = k[α 1,, α n ]. Henceforth we identify R en = k[x 1,, x n, y 1,, y n ] and R = R en /(x 1 y 1,, x n y n ). We have a Koszul complex where 0 Λ n (R en ) n ( 2n) Λ 2 (R en ) n ( 4) (R en ) n ( 2) R en 0 = (f 1,, f n ) (x 1 y 1 )f 1 + (x n y n )f n. It gives a projective resolution of R as an R en -module. Therefore for any R en -module M, HH(M) is the homology of 0 Λ n (R en ) n R en M( 2n) Λ 2 (R en ) n R en M( 4) M n ( 2) M 0. 2.2. Hochschild homology of Rouquier complexes. Now let w be a word with alphabet S S 1 and consider F w. If l(w) = r, then F w consists of r + 1 terms, i.e. By applying HH on this sequence, we have F w = 0 F r 1 w F 0 w 0 HH(F w ) = 0 HH(R, F r 1 w ) HH(R, F 0 w) 0 where each HH(R, F i w) is a bigraded k-vector space, where the bigrading comes from (1) the graded R-bimodule structure of F i w and (2) the grading of Hochschild homology. Now we define HHH(F w ) as the cohomology of this sequence. This is then a triply-graded k-vector space with (bimodule grading, Hochschild grading, cohomological grading). This turns out to be strongly related to the geometry of the knot (or link in general) represented by w as the following theorem shows. Theorem 2.1. [Kho07, Theorem 1] Up to a grading shift, HHH(F w ) only depends on the closure of the braid represented by w, say σ. This homology theory is isomorphic to the reduced link homology H(σ) defined by Khovanov and Rozansky (up to an affine transformation on the tri-grading.)

ROUQUIER COMPLEXES 9 One can check that H and HHH are defined in an analogous manner. The proof of the second part in [Kho07] uses this similarity. The first part follows from this result and the fact that H only depends on the closure of the braid σ. Indeed, we already know that Rouquier complexes only depend on the image of the word w in B W. However, if we take the closure, then we also need to consider the following Markov moves, which result in the same link if one takes the closure of braids. It is easy to prove the first part concerning the first Markov move; it suffices to show that HHH(A F s ) HHH(F s A ) for any A K b (SBim gr ) and s S. But we have A F s = A i 1 B s A i 2 (1) A i B s A i 1 (1) A i+1 B s A i (1) F s A = B s A i 1 A i 2 (1) B s A i A i 1 (1) B s A i+1 A i (1) Now the result follows from the fact that (MN) R (NM) R canonically (i.e. the coinvariants of MN and NM are isomorphic). 2.3. Connection with HOMFLY-PT polynomials. For each link σ, one can attach a unique Laurent polynomial P (σ) = P (σ)(a, z) with certain properties, called the HOMFLY-PT polynomial [FYH + 85]. It is characterized by the following properties. (1) P (unknot) = 1 (2) ap (L + ) a 1 P (L ) = zp (L 0 ) for links L +, L, L 0 that are isomorphic to one another except one part, which is described as follows. Jones polynomials and Alexander polynomials are the specializations of HOMFLY-PT polynomials: P (t 1, t 1/2 t 1/2 ), P (1, t 1/2 t 1/2 ), respectively. One property of H is that its Euler characteristic gives the HOMFLY-PT polynomial of the corresponding link. Here, we deal with a simple example. (This is the dual version of [Kho07, pp.15 16].) Suppose n = 1, thus R = k[α]. We also identify R en = k[x, y]. Let S = {s}. We calculate the HOMFLY-PT polynomial corresponding to s m B W. We assume that m is odd, so that the closure of the braid is a knot, not just a link.

ROUQUIER COMPLEXES 10 From Exercise 1.3, F s F s (m terms) is equivalent to the following chain complex. d B s (1 m) 0 d B s (3 m) 1 d m 2 B s (m 1) m R(m) We take HH on this sequence. First, the Koszul complex of R is very simple in this case. 0 k[x, y]( 2) 1 x y k[x, y] 0 Tensoring with R over R en, we have 0 R( 2) 0 R 0, thus HH 0 (R, R) = R, HH 1 (R, R) = R( 2), HH i (R, R) = 0 otherwise. If we tensor with B s over R en, then we have 0 B s ( 2) 1 1 α 1 1 α B s 0, thus HH 0 (R, B s ) = R(1), HH 1 (R, B s ) = R( 3), HH i (R, B s ) = 0 otherwise. Therefore, the zeroth and the first homology of HHH(F s F s ) is HH 0(d 0) HH 0(d 1) HH 0(d m 2) Id R(2 m) R(4 m) R(m) R(m) where HH 1(d 0) HH 1(d 1) HH 1(d m 2) 2α R( 2 m) R( m) R(m 4) R(m 2) HH 0 (d m 2 ), HH 0 (d m 4 ), = 0, HH 0 (d m 3 ), HH 0 (d m 5 ), = 1 2α, HH 1 (d m 2 ), HH 1 (d m 4 ), = 0, HH 1 (d m 3 ), HH 1 (d m 5 ), = 1 2α. Thus, we have nontrivial homology isomorphic to k at (m 4, 0, 1), (m 8, 0, 3),, (2 m, 0, m 2), (m, 1, 1), (m 4, 1, 3),, (2 m, 1, m). (Recall that the trigrading is given by (bimodule grading, Hochschild grading, cohomological grading).) Therefore, if we let then it follows that P m (x, y, z) := i,j,k x i y j z k dim HHH (i,j,k) (F s F s ), ( P m (x, y, z) = x m zx ( 2m x 4 y + 1 ) x 2 z ( m yz 2 + 1 )) x 4 z 2. Thus the Euler characteristic of HHH with ( respect to the second grading is P m (x, 1, z) = x m x ( 2 z 2 1 ) z m ( x 4 1 ) zx 2m) x 4 z 2. We put x x 1/2 z 1/4 and z x 1 z 1/2, and multiply x 3m 2 z m 4 to get f(m) := ( x 3m 2 z m 4 )Pm (x 1/2 z 1/4, 1, x 1 z 1/2 ) = xm 1 z 1 m 2 ( 1 + z m+1 + x 2 (z z m ))) z 2. 1 For example, we have f(1) = 1, f(3) = x 4 + x 2 z + x2 z, f(5) = x6 z x6 z + x4 z 2 + x4 z 2 + x4 f(7) = x 8 z 2 x8 z 2 x8 + x 6 z 3 + x6 z 3 + x6 z + x6 z

ROUQUIER COMPLEXES 11 We claim that these are HOMFLY-PT polynomials. Indeed, if we let g(m) be the usual HOMFLY- PT polynomial of the knot (m, 1), then from [kno] we have g(1) = 1, g(3) = a 4 + a 2 z 2 + 2a 2, g(5) = a 6 z 2 2a 6 + a 4 z 4 + 4a 4 z 2 + 3a 4, g(7) = a 8 z 4 4a 8 z 2 3a 8 + a 6 z 2 + 6a 6 z 4 + 10a 6 z 2 + 4a 6 If we substitute a x and z z 1/2 z 1/2, then they become the same as f(m). References [BT11] Brav, C. and Thomas, H., Braid groups and Kleinian singularities, Math. Ann. 351 (2011), no. 4, 1005 1017. [EK10] Elias, B. and Krasner, D., Rouquier complexes are functorial over braid cobordisms, Homology Homotopy Appl. 12 (2010), no. 2, 109 146. [EW] Elias, B. and Williamson, G., Lecture notes for Soergel bimodules and Kazhdan-Lusztig conjectures, Available at http://people.mpim-bonn.mpg.de/geordie/aarhus/. [FYH + 85] Freyd, P. J., Yetter, D. N., Hoste, J., Lickorish, W. B. R., Millett, K. C., and Ocneanu, A., A new polynomial invariant of knots and links, Bull. Amer. Math. Soc. 12 (1985), 239 246. [Jen17] Jensen, L. T., The 2-braid group and Garside normal form, Math. Z. 286 (2017), no. 1-2, 491 520. [Kho07] Khovanov, M., Triply-graded link homology and Hochschild homology of Soergel bimodules, Internat. J. Math. 18 (2007), no. 8, 869 885. [kno] The Knot Atlas, Available at http://katlas.org/wiki/. [KS02] Khovanov, M. and Seidel, P., Quivers, Floer cohomology, and braid group actions, J. Amer. Math. Soc. 15 (2002), 203 271. [Rou04] Rouquier, R., Categorification of the braid groups, Available at http://arxiv.org/abs/math/0409593, 2004. [Wei95] Weibel, C., An Introduction to Homological Algebra, no. 38, Cambridge University Press, 1995. Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139-4307, U.S.A. E-mail address: sylvaner@math.mit.edu