Higher representation theory in algebra and geometry: Lecture II

Higher representation theory in algebra and geometry: Lecture II Ben Webster UVA ebruary 10, 2014 Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 1 / 35

References or this lecture, useful references include: Chuang and Rouquier, Derived equivalences for symmetric groups and sl 2 -categorification (introduces notion of strong sl 2 categorification and covers several examples, such as the symmetric group) Lauda, A categorification of quantum sl(2) and Categorified quantum sl(2) and equivariant cohomology of iterated flag varieties (discusses the relationship with the cohomology of Grassmannians and is the source of the pictures) The slides for the talk are on my webpage at: http://people.virginia.edu/ btw4e/lecture-2.pdf Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 2 / 35

Reminder about last time So, last time we discussed the Lia algebra sl 2, its quantized universal enveloping algebra U q (sl 2 ) and how we might categorify them. The underlying idea was to look at these objects in a way that makes the categorification of them very natural, in connection with Grassmannians. In particular, the generators E and were connected to the diagram π Gr(k k + 1, S) 1 π 2 Gr(k, S) Gr(k + 1, S) We first interpreted this in terms of counting points, but then decided we could upgrade from this to considering cohomology or push and pull for sheaves on this diagram. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 3 / 35

Cohomology and bimodules H k := H (Gr(k, S)) k 1H k := H (Gr(k 1 k, S))(k 1) k+1h k := H (Gr(k + 1 k, S))(d k 1) E k (M) := k 1 H k Hk M k (M) := k+1 H k Hk M. Theorem E k+1 k = k 1 E k id [d 2k]q E k+1 k id [2k d]q = k 1 E k (k d/2) (k d/2) So, this is the relations of (quantum) sl 2 in a categorified form. But as a definition, we decided this was unsatisfying; we wouldn t know if we found other functors satsifying the same isomorphisms if we were really seeing the same thing. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 4 / 35

Natural transformations So, the thing that s missing is some understanding of what these isomorphisms are, but more generally of maps between functors. There s an obvious set of questions we haven t looked at at all: If I have two monomials in the functors and E, what are the natural transformations between them? If you don t like natural transformations, you can just think of these as maps between the tensor products of bimodules. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 5 / 35

Cohomology of Grassmannians and partial flag varieties If we want to get our hands dirty, we ll have to actually think a bit about the cohomology of Grassmannians. Of course, the Grassmannian carries a tautological vector bundle T Gr(k, S) S whose fiber over a point is the vector space itself; there s also the quotient S/T of the trivial bundle with fiber S by T. This gives us a bunch of cohomology classes, by taking the Chern classes of these bundles x i = c i (T) w i = c i (S/T). By the Whitney sum formula, these satisfy the relation (1+c 1 (T)t+c 2 (T)t 2 + )(1+c 1 (S/T)t+c 2 (S/T)t 2 + ) = 1+c 1 (S)t+. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 6 / 35

Cohomology of Grassmannians and partial flag varieties If we want to get our hands dirty, we ll have to actually think a bit about the cohomology of Grassmannians. Of course, the Grassmannian carries a tautological vector bundle T Gr(k, S) S whose fiber over a point is the vector space itself; there s also the quotient S/T of the trivial bundle with fiber S by T. This gives us a bunch of cohomology classes, by taking the Chern classes of these bundles x i = c i (T) w i = c i (S/T). By the Whitney sum formula, these satisfy the relation (1 + x 1 t + x 2 t 2 + )(1 + w 1 t + w 2 t 2 + ) = 1. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 6 / 35

Cohomology of Grassmannians and partial flag varieties So, for example: 1 = 1 You can solve for x i s in terms of w i s or vice versa. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 7 / 35

Cohomology of Grassmannians and partial flag varieties So, for example: x 1 + w 1 = 0 You can solve for x i s in terms of w i s or vice versa. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 7 / 35

Cohomology of Grassmannians and partial flag varieties So, for example: x 2 + x 1 w 1 + w 2 = 0 You can solve for x i s in terms of w i s or vice versa. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 7 / 35

Cohomology of Grassmannians and partial flag varieties So, for example: x 3 + x 2 w 1 + x 1 w 2 + w 3 = 0 You can solve for x i s in terms of w i s or vice versa. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 7 / 35

Cohomology of Grassmannians and partial flag varieties So, for example: x 3 + x 2 w 1 + x 1 w 2 + w 3 = 0 You can solve for x i s in terms of w i s or vice versa. Theorem The cohomology H (Gr(k, S)) of the Grassmannian is the quotient of the polynomial ring C[x 1,..., x k, w 1,..., w d k ] by these Whitney sum relations. Alternatively, you can consider an infinite polynomial ring C[x 1,..., w 1,... ] modulo the Whitney sum relation, and write all the cohomologies of all the Grassmannians as quotients of this by setting x i = 0 (i > k) w j = 0 (j > d k). Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 7 / 35

Natural transformations This also handled the question of one set of natural transformations: Proposition The natural transformations of the identity functor id k are the ring H (Gr(k, S)) itself. OK, that was easy. What about E k or k? The ring H (Gr(k k + 1, S)) can be found in a very similar way. But now, we have k and k + 1 dimensional tautological bundles T k, T k+1, and a line bundle T k+1 /T k. The notation can get confusing, since I have a class I was calling x i on H (Gr(k, S)) and on H (Gr(k + 1, S)), and these don t pullback to the same place: one goes to c i (T k ) and the other to c i (T k+1 ). Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 8 / 35

Cohomology and symmetric polynomials In the interest of not going crazy, let s look at this in a different way: let z 1,..., z k be the Chern roots of T k, and z k+1,..., z d be the Chern roots of S/T k. or us these are just formal symbols, which satisfy e i (1, k) = e i (z 1,..., z k ) = c i (T k ) e i (k + 1, d) = e i (z k+1,..., z d ) = c i (S/T k ) In the formulation with Chern roots, the Whitney sum relations simply say that the Chern roots of S are the union of those for T k and S/T k. The rest follows from the fact that: e i (1, d) = e i (1, k) + e i 1 (1, k)e 1 (k + 1, d) + + e i (k + 1, d). The Whitney sum relations for the Grassmannian simply say that e i (1, d) = 0 for all i > 0. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 9 / 35

Cohomology and symmetric polynomials Definition The coinvariant algebra C d is the quotient of C[z 1,..., z d ] by the relations e i (1, d) = 0. This inherits an S d action permuting the variables. Let S m1,...,m p := S m1 S mp S d, and C k 1,...,k p d Proposition = C S k 1,k 2 k 1,k 3 k 2,...,d kp d. The algebra H (Gr(k, S)) is isomorphic to Cd k which is generated by the elementary symmetric functions e i (1, k) and e i (k + 1, d). This can be generalized to the partial flag variety Gr(k 1 k 2 k p ). Proposition The cohomology Gr(k 1 k 2 k p ) is isomorphic to C k 1,k 2,...,k p d, generated by e i (1, k 1 ), e i (k 1 + 1, k 2 ),..., e i (k p + 1, d) in C d. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 10 / 35

Cohomology and symmetric polynomials Thus, the bimodule k+1 H k can be thought of as C k,k+1 d, with the left action by H k+1 = C k+1 d and the right action by H k = Cd k. Proposition Since H k+1 and H k generate C k,k+1 d, the bimodule maps from k+1 H k to itself are given by multiplication by elements of C k,k+1 d. Visually, we can represent these elements using diagrams: x i w j The left/right sides correspond to the left/right action, the bottom to the source bimodule and the top to the target. I represent the action of z k+1 by a dot on the line. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 11 / 35

Diagrams These diagrams have the huge advantage that I don t confuse the left and right actions. I can write a relation like c i (T k+1 ) = c i (T k ) + c 1 (T k+1 /T k )c i (T k ): x i = x i + x i 1 without making a mess separating the left and right actions. In general, I can describe any element of a tensor product of k±1 H k s using a diagram like this, with a line for each term in the tensor product. x i w j E Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 12 / 35

More natural transformations This describes all elements, but not all bimodule maps. or example, if d = 2, then 2 H 1 1 H 0 = C[t]/(t 2 ) C[t]/(t 2 ) C[t]/(t 2 ) = C 2, so we must get the matrices acting on here somehow. In general, k+2 H k+1 k+1 H k is always two copies of the same module. Gr(k k + 1 k + 2, S) π 1,3 π 1 π 2 Gr(k, S) Gr(k k + 2, S) Gr(k + 2, S) The map π 1,3 is a fiber bundle with fiber P 1. Thus as a bimodule: k+2h k+1 k+1 H k = H (Gr(k k + 1 k + 2, S)) = H (Gr(k k + 2, S)) 2 Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 13 / 35

Demazure operators rom the perspective of polynomials, this is also easy to see: if I consider the invariants of C k,k+1,k+2 d, this has a natural action of (k + 1, k + 2). Proposition The invariants and anti-invariants of (k + 1, k + 2) are both free modules of rank 1 over C k,k+2 d = H (Gr(k k + 2, S)). Thus, my bimodule maps as a ring are a 2 2 matrix algebra over C k,k+2 d. Definition Let the Demazure operator D i : C d C d be the map D i (f ) = f (z 1,..., z i+1, z i,..., z d ) f (z 1,..., z d ) z i+1 z i This kills polynomials invariant under (i, i + 1), and thus is well-defined. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 14 / 35

Demazure operators Geometrically, I can think of the Demazure operator as integration along the projection map Gr(k k + 1 k + 2, S) Gr(k k + 2, S). Important facts: D i (f ) = 0 if f = f (i,i+1) D i (fg) = D i (f )g + f (i,i+1) D i (g). Thus, D i is a module map for C i 1,i+1 d. D i (fg) = fd i (g) if f = f (i,i+1) D i (x i f ) x i+1 D i (f ) = x i D(f ) D i (x i+1 f ) = f, since D i (x i ) = 1, D i (x i+1 ) = 1. D i is the unique map satisfying these properties. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 15 / 35

Demazure operators Theorem Multiplication by C k,k+1,k+2 d and the Demazure operator D k+1 generate bimodule endomorphisms of k+2 H k+1 k+1 H k = C k,k+1,k+2 d. In terms of pictures, I ll draw a Demazure operator as a crossing: My relations then become: = Theorem These relations give a presentation of End( k+2 H k+1 k+1 H k ). Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 16 / 35

Demazure operators Theorem Multiplication by C k,k+1,k+2 d and the Demazure operator D k+1 generate bimodule endomorphisms of k+2 H k+1 k+1 H k = C k,k+1,k+2 d. In terms of pictures, I ll draw a Demazure operator as a crossing: My relations then become: = Theorem These relations give a presentation of End( k+2 H k+1 k+1 H k ). Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 16 / 35

Demazure operators OK, let s keep going; what about the triple tensor product? This is the endomorphisms of the module C k,k+1,k+2,k+3 d. Here, the Demazure operators D k+1 and D k+2 both commute with C k,k+3 d, and satisfy the relation: D k+1 D k+2 D k+1 (f ) = D k+2 D k+1 D k+2 (f ) = Theorem The endomorphisms of the functor k+l k are generated by the Demazure operators and multiplication by elements of C k,k+1,...,k+l d ; the relations these satsify are the ones we know. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 17 / 35

The nilhecke algebra Thus, on k+l k, there s a natural algebra that acts: Definition The nilhecke algebra N l is the algebra generated by symbols ψ 1,..., ψ l 1, y 1,..., y l subject to the relations: ψ i y i y i+1 ψ i = y i ψ i ψ i y i+1 = 1 ψ i ψ i+1 ψ i = ψ i+1 ψ i ψ i+1 ψ i ψ j = ψ j ψ i ( i j 1) y i y j = y j y i y j ψ j Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 18 / 35

The nilhecke algebra The nilhecke algebra has a very natural geometric realization: let l(l) be the l-step complete flag variety. We have 3 projections π 1,2, π 1,3, π 2,3 : l(l) 3 l(l) 2. We can endow H GL l (l(l) l(l)) with a convolution multiplication by Proposition A B = (π 1,3 ) (π 1,2A π 2,3B). We have a canonical isomorphism between the nilhecke algebra and (l(l) l(l)) endowed with the convolution multiplication. H GL l The polynomial generators y i come from the equivariant homology of l(l) ; the classes ψ i correspond to the fiber product l(l) Gr(1 i 1 i+1 l) l(l) l(l) l(l). Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 19 / 35

The nilhecke algebra The other way to think about the nilhecke algebra N l is its natural representation on k[z 1,..., z l ] via multiplication and Demazure operators Theorem This action induces an isomorphism y i f = z i f ψ i f = D i (f ). N l = Endk[z1,...,z l ] S l (k[z 1,..., z l ]) = M l! (k[z 1,..., z l ] S l ). In particular, N l and k[z 1,..., z l ] S l are Morita equivalent, and N l has a unique simple module C l which is l!-dimensional. It s useful for us to note that e l = ψ 1 ψ 2 ψ 1 ψ 3 ψ 2 ψ 1 ψ l ψ l 1 ψ 1 y l 1 1 y l 2 2 y l 1 is a primitive idempotent. Thus, for any N l module, M = (e l M) l!. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 20 / 35

Symmetric groups ix a field k, and consider the group algebras of the symmetric groups S n over k. These are related by induction and restriction functors. Let X m = i<m (im) by the mth Jucys-Murphy element. Proposition The action of X m on Res Sm S m 1 M commutes with the S m 1 -action. That is, it is a natural transformation of the functor Res Sm S m 1. ix a k, and let E a (M) = {h Res Sm S m 1 M (X m a) N h = 0 for N 0}. a (M) = {s h Ind Sm S m 1 M s(x m a) N h = 0 for N 0}. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 21 / 35

Symmetric groups In this case, 2 a carries an action of N 2 : there are natural transformations given by s n+1, X n+1 and X n+2. We can use these to define the operators ψ 1, y 1 and y 2 by the formulae: y 1 = X n+1 a y 2 = X n+2 a ψ 1 = (s n+1 + 1)(1 + (X i+1 X i ) + (X i+1 X i ) 2 + ) You can have some fun checking the relations. This extends to an action of N l on l a. OK, so that s at least two cases where a nilhecke action comes in. So maybe this is a universal feature of good sl 2 -actions? Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 22 / 35

Strong sl 2 -actions Definition A strong sl 2 -action on an additive category C consists of: A direct sum decomposition C = n Z C n. Adjoint functors such that satisfy the relations on C n : C n+2 C n E: C n C n+2 E = E id n (n 0) E id n = E (n 0) Compatible actions of the nilhecke algebra N l on l for every n, l, induced by nilpotent operators y: and ψ : 2 2. I m also going to assume integrability: or every object M, we have l M = E l M = 0 for l 0. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 23 / 35

Strong sl 2 -actions Examples: C d 2k = H k -mod: we ve shown all the properties except the adjunction of E and ; this is the statement that Hom Hk ( k±1 H k, H k ) = k H k±1. This map comes from relative Poincaré duality for the map Gr(k k + 1, S) Gr(k, S). n C n = m ks m -mod with functors (E a, a ): We ve checked all the conditions here, except [E, ] equation (and I never told you how to weight things)... A mixed blessing of this area is that the definitions are very flexible. There are lots of different ways of saying them, so you can weaken them when trying do something has a strong action, or strengthen them when trying get consequences. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 24 / 35

Strong sl 2 -actions Theorem (Chuang-Rouquier) Assume C is abelian and E and exact. Then the [E, ] relations hold if for each simple L, [L : EM] = [L : EM] + n[l : M]. for some set of objects M C n which you can filter any projective with. Thus, one can check this equation for M simple, but there are other interesting possibilities, like Specht modules for symmetric groups. The Specht modules S λ are a collection of representations of ZS n indexed by partitions of m. If char(k) = 0, then k Z S λ are a complete set of irreducible modules; in general, every projective ks m module has a Specht filtration. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 25 / 35

Symmetric groups Given a partition λ, we fill its box at (i, j) with its content j i k. The weight of a Specht module for (E a, a ) is #{boxes of content a ± 1} 2#{boxes of content a} + δ a,0 So if k = 3 and a = 0, then the partition (3, 1) has weight 2: 2 0 1 2 The module E a (k Z S λ ) (resp. a (k Z S λ )) has a filtration whose subquotients are k Z S µ where µ is λ minus (resp. plus) a box of content a. The [E, ] relations follow from counting Specht multiplicities on both sides of the equation. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 26 / 35

Symmetric groups Given a partition λ, we fill its box at (i, j) with its content j i k. The weight of a Specht module for (E a, a ) is #{boxes of content a ± 1} 2#{boxes of content a} + δ a,0 So if k = 3 and a = 1, then the partition (3, 1) has weight 1: 2 0 1 2 The module E a (k Z S λ ) (resp. a (k Z S λ )) has a filtration whose subquotients are k Z S µ where µ is λ minus (resp. plus) a box of content a. The [E, ] relations follow from counting Specht multiplicities on both sides of the equation. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 26 / 35

Symmetric groups Given a partition λ, we fill its box at (i, j) with its content j i k. The weight of a Specht module for (E a, a ) is #{boxes of content a ± 1} 2#{boxes of content a} + δ a,0 So if k = 3 and a = 2, then the partition (3, 1) has weight 2: 2 0 1 2 The module E a (k Z S λ ) (resp. a (k Z S λ )) has a filtration whose subquotients are k Z S µ where µ is λ minus (resp. plus) a box of content a. The [E, ] relations follow from counting Specht multiplicities on both sides of the equation. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 26 / 35

Representations of gl n Let B be an Artinian category of representations of gl m (or even the superalgebra gl m q ) which is closed under tensor product with finite dimensional modules (finite dimensionals, parabolic category O, etc.). Definition Let Ω = e i,j e j,i gl n gl n. E a (M) := {g M C m (Ω a) p g = 0 for p 0} a (M) := {g M (C m ) (Ω a) p g = 0 for p 0} This has a N l -action on l a much like that for symmetric groups: y = Ω a ψ = (1 s)(1 Ω 13 + Ω 2 13 ) Theorem The category B has a strong categorical sl 2 action by E a, a ; the weight is a function of how higher Casimirs act on a module. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 27 / 35

Divided powers I hope you re now somewhat convinced that this isn t a totally vacuous notion. But it s not a useful one if it doesn t have consequences. Assume from now on that C = n C n is a category with strong sl 2 action by functors E, : C C. The first consequence comes from the fact that N l is a matrix algebra. Definition Let E (l) = e l E l and (l) = e l l. We have that E l = (E (l) ) l! l = ( (l) ) l! Thus, these functors categorify the divided powers E (l) := E l /(l!) and (l) := l /(l!). Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 28 / 35

Uniqueness We have a commutation formula E (l) (l) = l (j) ((H 2j)(H 2j 1)... (H l j + 1) E (j). (l j)! j=0 In particular, if v is a highest weight vector of weight d, then E (d) (d) v = H(H 1) (H d + 1) v = v. d! Thus, if M is an object in C d such that EM = 0, then E (d) (d) M = M. Proposition In this case, we have a natural isomorphism End(M) = End( (d) M) Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 29 / 35

Uniqueness These divided powers help us to work out the structure of a categorification. Theorem Assume that M is an object in C d such that End(M) = k and EM = 0. Then End( (k) M) = H k = H (Gr(k, C d ); k); in particular, (k) M is indecomposable. Steps in the proof: Note that End( k M) End( d M) = End k (C d ) must be injective. This shows that no class in k[y 1,..., y k ] S k that goes to 0 in End( k M) can go to 0 in H (Gr(k, C d ); k). On the other hand, ( ) d dim End( (k) M) = dim Hom(M, E (k) (k) M) =. k This is only possible if the map k[y 1,..., y k ] S k End( (k) M) induces the desired isomorphism. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 30 / 35

Uniqueness This is one version of a uniqueness theorem: Theorem (Chuang-Rouquier) If C has Grothendieck group isomorphic to V d = Sym d 1 (C 2 ), and C d = k -mod, then Cd 2k is equivariantly equivalent to H k -proj. You might rightly protest that you might have C d k -mod; this is an issue that s easily fixed, though: We have a natural map from k[y 1,..., y d ] S d End( (d) M) = End(M); the more general version of this theorem is that End( (k) M) = k[y 1,..., y d ] S k,d k k[y1,...,y d ] S d End(M). Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 31 / 35

Next time... So, next time, we ll talk about 2-categories, how to fix the [E, ] relation, and other issues of tightening the definition, and discuss the marquee application of this theory: the proof of the Broué conjecture for symmetric groups. Like last time, we ve get up the pieces set up on the board, but we have to wait until next time to take the king. Ben Webster (UVA) HRT : Lecture II ebruary 10, 2014 32 / 35