Diagram categories for U q -tilting modules at q l = 1

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1 Diagram categories for U q -tilting modules at q l = 1 Or: fun with diagrams! Daniel Tubbenhauer s t s t s s s t s t s s = s s s t t s s s s t t s Joint work with Henning Haahr Andersen October 2014 Daniel Tubbenhauer October / 30

2 1 The why of diagram categories String calculus Biadjoint functors 2 Categorification of Hecke algebras Hecke algebras and Soergel bimodules Soergel s categorification 3 Let us use diagrams! The i are selfadjoint functors Diagrammatic categorification 4 What about U q -modules at roots of unity? Daniel Tubbenhauer October / 30

3 String calculus for 2-categories - Part 1 Question: Can we interpret Cat 2 using diagrams? Let us start with Cat 1 : Instead of C 1 D use the Poincaré dual D C Composition becomes E Not really spectacular... D 2 E C 2 D D 1 C 1 = D = C 2 1 E 2 1 E D C Daniel Tubbenhauer String calculus October / 30

4 String calculus for 2-categories - Part 2 Let us go to Cat 2 now: Think of a natural transformations α,β, as a proceeding in time: 2 1 D D α C = α: E β C = β: We do not draw identities 1 id D C = id: C γ D = γ: H 2 H 1 id H 2 H 1 Daniel Tubbenhauer String calculus October / 30

5 String calculus for 2-categories - Part 3 Compositions? Sure! Vertical: E β C v E D β D β C = E D C β and horizontal E α D h D α C = E α D α C That looks promising: 2-categories are like 2-dimensional spaces. Daniel Tubbenhauer String calculus October / 30

6 Adjoint functors abstract Definition(Dan Kan 1958) Two functors : C D and : D C are adjoint iff there exist natural transformations called unit ι: id C and counit ε: id D such that id ι ε id and ι id id ε id id commute. Here is the left adjoint of. Example forget: Q-Vect Set has a left adjoint free: Set Q-Vect. In words: If you have lost your key, then the only guaranteed solution is to search everywhere. Daniel Tubbenhauer Biadjoint functors October / 30

7 Adjoint functors such that I understand Let us draw sting pictures! D id D id = D C, id = C D, ι = C and ε = D C id C Adjointness is just straightening of the strings C D = D C and D C = C D Daniel Tubbenhauer Biadjoint functors October / 30

8 Biadjoint functors = Isotopies If is also the right adjoint of, then the picture gets topological. Biadjointness is just straightening of the strings! irst left C D = D C and D C = C D then right C D = D C and D C = C D Daniel Tubbenhauer Biadjoint functors October / 30

9 Biadjoint functors in nature (do not ask me for details...) Categories of modules over finite dimensional symmetric algebras and their derived counterparts have plenty of built-in biadjoint functors (tensoring with certain bimodules). Prominent examples are finite groups and induction and restriction functors between them. Various categories arising in representation theory of Hecke algebras and category O admit lots of biadjoint functor. or example translation functors and Zuckerman functors. Every (extended) TQT : Cob n+2 Vec 2 gives a bunch of biadjoint functors: ((M),(τ(M))) for any n+1 manifold M where τ flips M. Prominent examples come from commutative robenius algebras for n = 2, Witten-Reshetikhin-Turaev TQT s for n = 3, Donaldson-loer for n = 4, and way more... Other fancy stuff like ukaya-loer categories, derived categories of constructible sheaves on flag varieties... Daniel Tubbenhauer Biadjoint functors October / 30

10 The Iwahori-Hecke algebra (for me n = 3 is enough) Let us fix n = 3. Then the group ring of the symmetric group Q[S 3 ] has two generators s 1,s 2. They satisfy s 2 1 = 1 = s 2 2 and s 1 s 2 s 1 = s 2 s 1 s 2. Iwahori: The Hecke algebra H 3 = H[S 3 ] is a q-deformation of Q[S 3 ]. Definition/Theorem(Iwahori 1965) The Hecke algebra H 3 has generators T 1,T 2 and relations T 2 1 = (q 1)T 1,2 +q = T 2 2 and T 1 T 2 T 1 = T 2 T 1 T 2. The classical limit q 1 gives Q[S 3 ]. Nowadays Hecke algebras à la Iwahori appear everywhere, e.g. low dimensional topology, combinatorics, representation theory of gl n etc. Daniel Tubbenhauer Hecke algebras and Soergel bimodules October / 30

11 Idempotents are better! Recall that primitive idempotents e j A in any finite dimensional Q-algebra A give rise to Ae j which is indecomposable. The group algebra Q[S 3 ] admits idempotents : i 1 = 1+s 1 and i 2 = 1+s 2, because they satisfy i 2 1 = 2i 1,2 = i 2 2 and i 1 i 2 i 1 +i 2 = i 2 i 1 i 2 +i 1. or the Hecke algebra: Set t = q and define b 1,2 = t 1 (1+T 1,2 ) (we see the Hecke algebra over Q[t,t 1 ] now). The b 1,b 2 satisfy b 2 1 = (t +t 1 )b 1,2 = b 2 2 and b 1 b 2 b 1 +b 2 = b 2 b 1 b 2 +b 1. Only positive coefficients? Suspicious... Daniel Tubbenhauer Hecke algebras and Soergel bimodules October / 30

12 Bimodules do the job? Take R = Q[X 1,X 2,X 3 ] (with degree of X i = 2) and define the s 1,2 -invariants as R s1 = {p(x 1,X 2,X 3 ) R p(x 1,X 2,X 3 ) = p(x 2,X 1,X 3 )} and R s2 = {p(x 1,X 2,X 3 ) R p(x 1,X 2,X 3 ) = p(x 1,X 3,X 2 )}. or example X 1 +X 2 R s1, but X 1 +X 2 R s2. The algebra R is a (left and right) R s1,2 -module. Thus, B 1 = R R s 1 R{ 1} and B 2 = R R s 2 R{ 1} are R-bimodules. Write short B ij for B i R B j. unny observation (i = 1,2): B ii = Bi {+1} B i { 1} and B 121 B 2 = B212 B 1. We have seen this before... Daniel Tubbenhauer Hecke algebras and Soergel bimodules October / 30

13 The combinatoric of S 3 The bimodule world: Take tensor products B i of the B i s. The atoms of the bimodules world are the indecomposables: All M such that M = M 1 M 2 implies M 1,2 = 0. We have B = R, B 1 = B 1, B 2 = B 2, B 12 = B 12 and B 21 = B 21 as atoms, but B 121 = B1 R R S 3 R{ 3} and B 212 = B2 R R S 3 R{ 3} s 2 s 1 B 1 B 21 R s 1 s 2 B 121 B 12 B 2 s 2 s 1 and B 121 = B 212 = R R S 3 R{ 3} is indecomposable. There are exactly as many indecomposables as elements in S 3. Suspicious... Daniel Tubbenhauer Hecke algebras and Soergel bimodules October / 30

14 What is a morphism of elements of H 3? Definition(Soergel 1992) Define SC(3) to be the category with the following data: Objects are (shifted) direct sums of tensor products B i of B i s. Morphisms are matrices of (graded) bimodule maps. Theorem(Soergel 1992) SC(3) categorifies H 3. The indecomposables categorify the Kazhdan-Lusztig basis elements of H 3. Morally: SC(3) is the categorical analogon of H 3. The morphisms in SC(3) are invisible in H 3. Wait: What do you mean by categorify? Daniel Tubbenhauer Soergel s categorification October / 30

15 (Split) rothendieck group If you have a suitable category C, then we can easily collapse structure by totally forgetting the morphisms: The (split) rothendieck group K0 (C) of C has isomorphism classes [M] of objects M Ob(C) as elements together with [M 0 ] = [M 1 ]+[M 2 ] M 0 = M1 M 2,[M 1 ][M 2 ] = [M 1 M 2 ] and [M{s}] = t s [M]. This is a Z[t,t 1 ]-module. Example We have K 0 (Q-Vect gr) = Z[t,t 1 ], [Q{s}] t s 1. The whole power of linear algebra is forgotten by going to K 0 (Q-Vect) gr. Daniel Tubbenhauer Soergel s categorification October / 30

16 Categorification? We have two functors 1 = B 1 R and 2 = B 2 R. These are additive endofunctors of SC(3). Thus, the introduce an action [ i ] on K0 ( SC(3)). We have a commuting diagram (we ignore to tensor with Q(t)) K 0 ( SC(3)) [ i] K 0 ( SC(3)) φ = = φ H 3 b i H 3. Thus, the functors 1, 2 categorify the multiplication in H 3! Said otherwise: They categorify the action of H 3 on itself. Moreover, the indecomposables give a good basis of H 3. Daniel Tubbenhauer Soergel s categorification October / 30

17 The speaker is lost... The speaker is lost: That was too abstract. Can we understand this topological? Observation(Elias-Khovanov 2009) The functors 1 and 2 are selfadjoint! Thus, there is a stringy calculus for SC(3). As before: Well denote compositions like by SC(3) 1 SC(3) 2 SC(3) 2 SC(3) 1 SC(3) 1 SC(3) or simplified Think: Apply to R on the right. Daniel Tubbenhauer The i are selfadjoint functors October / 30

18 enerators We have the following one color generators: deg = +1 deg = +1 deg = 1 deg = 1 deg = 0 1 id id deg = +1 deg = +1 deg = 1 deg = 1 deg = 0 2 id id Daniel Tubbenhauer The i are selfadjoint functors October / 30

19 Exempli gratia deg = These Soergel diagrams can get very complicated, but this is an information completely invisible in H 3. Daniel Tubbenhauer The i are selfadjoint functors October / 30

20 Some relations - Part 1 We need some additional relations to make the story work. Some are combinatorial (which we do not recall), but, due to biadjointness, some are topological. = = = = = = Daniel Tubbenhauer The i are selfadjoint functors October / 30

21 Some relations - Part 2 Some are really topological: There is more than planar isotopies. The functors 1 and 2 are robenius. This gives = = = = = Daniel Tubbenhauer The i are selfadjoint functors October / 30

22 The diagram category suffices Definition(Elias-Khovanov 2009) Define DC(3) to be the category with the following data: Objects are (shifted) formal direct sums of sequences of the form Morphisms are matrices of (graded) Soergel diagrams module the local relations. Theorem(Elias-Khovanov 2009) There is an equivalence of graded, monoidal, Q-linear categories DC(3) = SC(3). Conclusion: The (seemingly very rigid) Hecke algebra H 3 has an overlying topological counterpart! Daniel Tubbenhauer Diagrammatic categorification October / 30

23 Some upshots of Elias-Khovanov s approach No restriction to S 3 : Any Coxeter system works. Diagrammatic categorification is low tech. Playing with diagrams is fun, easy and the topological flavour gives new insights. or example, Elias and Williamson s algebraic proof that the Kazhdan-Lusztig polynomials have positive coefficients for arbitrary Coxeter systems was discovered using the diagrammatic framework. New insights into topology: Elias used the topological behaviour to give a new categorification of the Temperley-Lieb algebra. Rouquier produced a braid group action on (chain complexes of) Soergel diagrams. This is functorial: It also talks about braid cobordisms (these live in dimension 4!). Rouquier s results can be extended to give HOMLY-PT homology. This still mysterious homology is related to knot loer homology. More is to be expected! Daniel Tubbenhauer Diagrammatic categorification October / 30

24 Non-associative=bad Recall that sl 2 is [, ]-spanned by = ( ) ( , E = 0 1 ) ( 0 0 and H = ). Non-associative: Take U( ): LieAlg AssoQ-Alg which is the left adjoint of [, ]: AssoQ-Alg LieAlg. Thus, the universal envelope U(sl 2 ) is the free, associative Q-algebra spanned by symbols E,,H,H 1 modulo HH 1 = H 1 H = 1, HE = EH and H = H. By magic: sl 2 -Mod = U(sl 2 )-Mod. E E = H. Naively quantize: U q (sl 2 ) = U q is the free, associative Q(q)-algebra spanned by symbols E, and K,K 1 (think: K = q H,K 1 = q H ) modulo KK 1 = K 1 K = 1, EK = q 2 KE and K = q 2 K. E E = K K 1 q q 1 (think: qh q H q q 1 q 1 H). Daniel Tubbenhauer October / 30

25 What are the atoms? act of life If q is an indeterminate, then U q has the same representation theory as sl 2. In particular, U q -Mod fin is semisimple: Atoms are the irreducibles. If q l = 1, then this totally fails: U q -Mod fin is far away to be semisimple. Why do we want to study something so nasty? Magic: Many similarities to the representation theory of a corresponding almost simple, simply connected algebraic group modulo p. Many similarities to the representation theory of a corresponding affine Kac-Moody algebra. It provides ribbon categories (link invariants) which can be semisimplified to provide modular categories (2 + 1-dimensional TQT s). It turns out that the right atoms are the so-called indecomposable U q -tilting modules. The corresponding category T is what we want to understand. Daniel Tubbenhauer October / 30

26 Translation functors Principle(Bernstein-elfand-elfand 1970) Do not study representations explicitly: That is too hard. Study the combinatorial and functorial behaviour of their module categories! So let us adopt the B principle from category O! In particular, there are two endofunctors Θ s,θ t of T λ (there is a decomposition of T into blocks T λ ) called translation through the s,t-wall. These are selfadjoint robenius functors with combinatorial behaviour governed by the -dihedral group D = {s,t s 2 = 1 = t 2 }: Θ s Θ s = Θs Θ s and Θ t Θ t = Θt Θ t. We have seen something similar before: There should be a diagram category (inspired by the corresponding one for H(D )) that governs T and pend(t). Daniel Tubbenhauer October / 30

27 Sorry: No tenure means I have to stress my own results Definition/Theorem(Elias 2013) There is a diagram category D( ) that categorifies H(D ) (that is what we are looking for!). The indecomposables categorify the Kazhdan-Lusztig basis elements of H(D ). Definition/Theorem There is are diagram categories QD( ) and Mat fs ( QD( )) c for T and pend(t). The diagram categories are naturally graded which introduce a non-trivial grading on T and pend(t). We have K 0 (Tgr λ ) = B : Thus, T gr λ categorifies the Burau representation B of the braid group B in -strands (cut-offs are possible). The action of B is categorified using certain chain complexes of truncations of Θ s,θ t. We have K0 (pend(tgr λ )) = TL q : Thus, pend(tgr λ ) categorifies (a certain summand of) the Temperley-Lieb algebra in -strands (cut-offs are possible). Daniel Tubbenhauer October / 30

28 Elias dihedral cathedral The category D( ) is almost as before, but easier: No relations among the colors red s and green t: Neither nor Pictures look like Our QD( ) looks similar plus some extra relations. Daniel Tubbenhauer October / 30

29 Open connections Question: What is the non-trivial grading (purely a root of unity phenomena) trying to tell us about the link and 3-manifold invariants deduced from T? Question: Similarly, what is the non-trivial grading (purely a root of unity phenomena) trying to tell us about algebraic groups modulo p? We argue that each block T gr λ separately can be used to obtain invariants of links and tangles - there are very explicit relations to (sutured) Khovanov homology and bordered loer homology. Hence, each block T gr λ separately yields information about link and tangle invariants in the non-root of unity case, while the ribbon/modular structure of T yields the Witten-Reshetikhin-Turaev invariants. Question: What is going on here? As in the H(S n ) case: Question: Is there a cobordism theory that explains the grading and the robenius structure topological? Daniel Tubbenhauer October / 30

30 There is still much to do... Daniel Tubbenhauer October / 30

31 Thanks for your attention! Daniel Tubbenhauer October / 30

Categorification, Lie algebras and Topology

Categorification, Lie algebras and Topology Ben Webster Northeastern University/University of Oregon June 17, 2011 Ben Webster (Northeastern/Oregon) Categorification, Lie algebras and Topology June 17,