higher representation theory in algebra and geometry ... et resume… · the construction of knot...

Higher representation theory in algebra and geometry:Lecture VIII

Ben Webster

April 8, 2014

Ben Webster (UVA) HRT : Lecture VIII April 8, 2014 1 / 36

References

For this lecture, useful references include:

B.W., Knot invariants and higher representation theory

The slides for the talk are on my webpage at:http://people.virginia.edu/~btw4e/lecture-8.pdf

You can also find some proofs that I didn’t feel like going through in class at:https://pages.shanti.virginia.edu/Higher_Rep_Theory/

The future

So, there are 3 class meetings left. In what time is left, I want to try to covertwo interesting applications of the theory we’ve discussed.

the construction of knot invariants using this theory. We’ve alreadydiscussed one special case of this, using sl2-categorifications to obtainthe Jones polynomial. This generalizes to other types. There’s also a“dual” construction of these knot invariants for sln, which we’ll likelyget to in Lecture 9. This also includes some interesting connections toalgebraic geometry.

the perspective on the representation theory of Cherednik algebrasafforded by higher representation theory. This is is, of course, anenormous topic, but I think it’s an exciting application of the theory, andone worth discussing a bit. I anticipate that will be Lecture 10.

Knot invariants

Roadmap

quantum groups Uq(g)

ribbon category of Uq(g)-reps

quantum knot polynomials(Jones polynomial, etc.)

Khovanov-Lauda/Rouquier2-categories U

quantum knot homologies

Knot invariants

Roadmap

Knot invariants

Roadmap

ribbon 2-category of U-reps?

Knot invariants

Roadmap

categorifications of tensorproducts of simples

Knot invariants Braiding

Reshetikhin-Turaev invariants

Let me briefly indicate how the left side of the diagram works.

Quantum groups are deformations of universal enveloping algebras. Perhapsthe most important thing about them is that they deform the tensor productof U(g) representations. Given two reps V , W, we still have a Uq(g)-action onV ⊗W.

However, in this new definition, the obvious map V ⊗W → W ⊗ V is not amap of representations. Luckily, this can be fixed by changing the map a littlebit, and multiplying by a formal sum R ∈ Uq(g)⊗ Uq(g) called the “universalR-matrix.”

bV,W : V ⊗W R·−→ V ⊗Wflip−→ W ⊗ V

Let me briefly indicate how the left side of the diagram works.

Quantum groups are deformations of universal enveloping algebras. Perhapsthe most important thing about them is that they deform the tensor productof U(g) representations. Given two reps V , W, we still have a Uq(g)-action onV ⊗W.

However, in this new definition, the obvious map V ⊗W → W ⊗ V is not amap of representations. Luckily, this can be fixed by changing the map a littlebit, and multiplying by a formal sum R ∈ Uq(g)⊗ Uq(g) called the “universalR-matrix.”

bV,W : V ⊗W R·−→ V ⊗Wflip−→ W ⊗ V

Proposition

The maps bV,W make Uq(g) into a braided monoidal category.

One way to think about this fact is that if you represent

bV,W 7→

Then the maps induced by switching factors of big tensor products satisfy thebraid relations.

(1W ⊗ bU,V)(bU,W ⊗ 1V)(1U ⊗ bV,W)

(bV,W ⊗ 1U)(1V ⊗ bU,W)(bU,V ⊗ 1W)

On the other hand bV,WbW,V 6= 1, as the picture above suggests.

Proposition

The maps bV,W make Uq(g) into a braided monoidal category.

One way to think about this fact is that if you represent

bV,W 7→

Then the maps induced by switching factors of big tensor products satisfy thebraid relations.

(1W ⊗ bU,V)(bU,W ⊗ 1V)(1U ⊗ bV,W)

(bV,W ⊗ 1U)(1V ⊗ bU,W)(bU,V ⊗ 1W)

On the other hand bV,WbW,V 6= 1, as the picture above suggests.

Knot invariants Cups and caps

The other important structure on representations of a quantum group is takingdual of representations. As with switching tensor factors, we have to becareful about left and right. There is a contravariant functor V 7→ V∗ calledright dual (there’s also left dual which is the same vector space with adifferent Uq(g)-action).

The category of Uq(g)-representations has canonical maps

evaluation V∗ ⊗ V → C(q), represented by

coevaluation C(q)→ V ⊗ V∗, represented by

If you want the maps the other way, you need to take left dual.

Not all is lost! After all, we have a map which switches tensor factors. Butshould we take

Of course, we can’t play favorites. Instead we should take the geometricmean.

If V is irreducible, there’s a unique constant aV ∈ C(q) (actually a power of q)such that

√aV .

A natural choice of√

aV (I really mean functorial) is called a ribbonstructure. The reason for the name is that if we interpret the diagrams asdrawn with ribbon, then they are with a left and right twist added,respectively.

DefinitionThis map is called quantum trace and its vertical flip is called quantumcotrace.

This allows us to associate a map for any oriented tangle labeled withrepresentations, by associating the braiding to a crossing and appropriate traceor evaluation to cups:

C[q, q−1]

V V V∗

V V∗

Composing these together for a given ribbon link results in a scalar: theReshetikhin-Turaev invariant for that labeling.

This allows us to associate a map for any oriented tangle labeled withrepresentations, by associating the braiding to a crossing and appropriate traceor evaluation to cups:

C[q, q−1]

V V V∗

V V∗

Composing these together for a given ribbon link results in a scalar: theReshetikhin-Turaev invariant for that labeling.

A historical interlude

Progress has been made on categorifying these in a piecemeal fashion for awhile

Khovanov (’99): Jones polynomial (C2 for sl2).

Oszvath-Szabo, Rasmussen (’02): Alexander polynomial (which isactually a gl(1|1) invariant, and doesn’t fit into our general picture).

Khovanov (’03): C3 for sl3.

Khovanov-Rozansky (’04): Cn for sln.

Stroppel-Mazorchuk, Sussan (’06-’07): ∧iCn for sln.

Cautis-Kamnitzer (’06): ∧iCn for sln.

Khovanov-Rozansky(’06): Cn for son.

What I’ll give you is a unified, pictorial construction that should include all ofthese. For that, we need tensor products. p=proven, c=conjectured.

p Khovanov (’99): Jones polynomial (C2 for sl2).

? Oszvath-Szabo, Rasmussen (’02): Alexander polynomial (which isactually a gl(1|1) invariant, and doesn’t fit into our general picture).

p Khovanov (’03): C3 for sl3.

p Khovanov-Rozansky (’04): Cn for sln.

p Stroppel-Mazorchuk, Sussan (’06-’07): ∧iCn for sln.

p Cautis-Kamnitzer (’06): ∧iCn for sln.

c Khovanov-Rozansky(’06): Cn for son.

p Khovanov (’99): Jones polynomial (C2 for sl2).

? Oszvath-Szabo, Rasmussen (’02): Alexander polynomial (which isactually a gl(1|1) invariant, and doesn’t fit into our general picture).

p Khovanov (’03): C3 for sl3.

p Khovanov-Rozansky (’04): Cn for sln.

p Stroppel-Mazorchuk, Sussan (’06-’07): ∧iCn for sln.

p Cautis-Kamnitzer (’06): ∧iCn for sln.

c Khovanov-Rozansky(’06): Cn for son.

Tensor products Definition

Tensor products

In the case of sl2, we introduced a graphical calculus for elements ofVλ = Vλ1 ⊗ · · · ⊗ Vλ` .

A downward black line on the left means acting by Fi.

A red line at the left labeled by λ corresponds to vλ ⊗−, where vλ is thehighest weight vector of Vλ.

So, we obtain a spanning set of Vλ consisting of vectors like

Fi(vλ1 ⊗ Fjvλ2)↔λ1 + λ2−αj + αi

λ1 + λ2−αjλ2 λ2 − αj

iλ1jλ2

Tensor products

In the case of sl2, we introduced a graphical calculus for elements ofVλ = Vλ1 ⊗ · · · ⊗ Vλ` .

A downward black line on the left means acting by Fi.

A red line at the left labeled by λ corresponds to vλ ⊗−, where vλ is thehighest weight vector of Vλ.

So, we obtain a spanning set of Vλ consisting of vectors like

Fi(vλ1 ⊗ Fjvλ2)↔λ1 + λ2−αj + αi

λ1 + λ2−αjλ2 λ2 − αj

iλ1jλ2

Tensor products

Let Tλ be the algebra whose elements are k-linear combinations of immersed1-manifolds with

black components oriented, dotted and labeled with i ∈ Γ andred components have no intersections, and are labeled with the weights λin order modulo the relations

λi ii λ

a+b=λi−1b

any diagram witha black line at

the far left is 0.

and. . . . . .Ben Webster (UVA) HRT : Lecture VIII April 8, 2014 12 / 36

Diagrams

unless i = j

Qij(y1, y2)

unless i = k = j± 1

Qij(y3, y2)− Qij(y1, y2)

y3 − y1

Diagrams

unless i = j

Qij(y1, y2)

unless i = k = j± 1

Qij(y3, y2)− Qij(y1, y2)

y3 − y1

Categorical action

Recall, last time, we defined the notion of a categorical action of g. For this,we need functors Fi and Ei.

These are induction and restriction functors, which can think of as tensorproduct with the bimodules:

Fi = · · ·

· · ·

right action

left action

E = · · ·

· · ·

right action

left action

The action of Rm on the power Fm is by attaching pictures at the bottom.Adjunction is essentially automatic.

The tricky part is checking the sl2 relations. This is hard.

Grothendieck groups

Theorem

The GG of Tλ -pmod is the Lusztig integral form of Vλ, sending the functor Fi

to the action of Fi, and the functor λ (adding a red line) to the inclusion

V−⊗vhigh↪→ V ⊗ Vλ.

But we’d like to talk about the category Tλ -mod, which doesn’t have thesame Grothendieck group: the map

K0(Tλ -pmod)→ K0(Tλ -mod)

is injective, but not surjective, since not all simple modules have finiteprojective resolutions. (Think about k[x]/(x2)).

However, this map is an isomorphism after tensoring with C(q), so everyfinite dimensional Tλ-module defines a class in Vλ.

Grothendieck groups

Theorem

The GG of Tλ -pmod is the Lusztig integral form of Vλ, sending the functor Fi

to the action of Fi, and the functor λ (adding a red line) to the inclusion

V−⊗vhigh↪→ V ⊗ Vλ.

But we’d like to talk about the category Tλ -mod, which doesn’t have thesame Grothendieck group: the map

K0(Tλ -pmod)→ K0(Tλ -mod)

is injective, but not surjective, since not all simple modules have finiteprojective resolutions. (Think about k[x]/(x2)).

However, this map is an isomorphism after tensoring with C(q), so everyfinite dimensional Tλ-module defines a class in Vλ.

What does the representation theory of this algebra look like?

Projectives are just summands of the modules Pκi = Tλe(i, κ) wheree(i, κ) is the sequence corresponding to a particular ordering of red andblack dots. The indecomposables give you a “canonical basis” (Lusztig’sif g is symmetric type).

Simple modules are endowed with a crystal structure (exactly as inLauda and Vazirani), which is the tensor product of the crystals for Vλi .These give you a “dual canonical basis.”

These objects both give bases of the Grothendieck group which are not verycompatible with the tensor product structure. If we’re ever going to do anycalculations, we’re going to need objects that correspond to pure tensors.

Tensor products Standard modules

Standard modules

Well, how would we construct the pure tensor v1 ⊗ Fiv2? We have modulescorresponding to

λ1 λ2 i

Fi(v1 ⊗ v2) = v1 ⊗ Fiv2 + qλiFiv1 ⊗ v2

λ1 λ2i

Fiv1 ⊗ v2

So we’d like to subtract the former from the latter. Of course, in categoriesyou can’t subtract, but you can look for submodules. As it happens, the mapgiven by is injective, so modding out by its image gives a module withthe right class in the Grothendieck group.

Can this phenomenon be generalized?

Standard modules

Well, how would we construct the pure tensor v1 ⊗ Fiv2? We have modulescorresponding to

λ1 λ2 i

Fi(v1 ⊗ v2) = v1 ⊗ Fiv2 + qλiFiv1 ⊗ v2

λ1 λ2i

Fiv1 ⊗ v2

So we’d like to subtract the former from the latter. Of course, in categoriesyou can’t subtract, but you can look for submodules. As it happens, the mapgiven by is injective, so modding out by its image gives a module withthe right class in the Grothendieck group.

Can this phenomenon be generalized?

Standard modules

a “left” crossing a “right” crossing

DefinitionThe standard module Sκλ is the quotient of Pκλ by the submodule generated byall diagrams with at least one “left” crossing as above, and no “right”crossings.

Put another way, we can associate a composition to the module Pκi bycounting the number of black strands between each pair of reds, and we modout by the images of all maps from projectives strictly higher in dominanceorder.

In the example of the last slide, we just use that (1, 0) > (0, 1).

Standard modules

a “left” crossing a “right” crossing

DefinitionThe standard module Sκλ is the quotient of Pκλ by the submodule generated byall diagrams with at least one “left” crossing as above, and no “right”crossings.

Put another way, we can associate a composition to the module Pκi bycounting the number of black strands between each pair of reds, and we modout by the images of all maps from projectives strictly higher in dominanceorder.

In the example of the last slide, we just use that (1, 0) > (0, 1).

Standard modules

As you may have guessed

Proposition

[Sκi ] = Fiκ(1)−1 · · ·Fi1v1 ⊗ · · · ⊗ Fin · · ·Fκ(`)vn

This makes standard modules invaluable as “test objects” for functors to seethat they behave correctly on the Grothendieck group.

For example, FiSκi has a filtration which categorifies the usual formula

∆(`)(Fi) = Fi ⊗ K̃i ⊗ · · · ⊗ K̃i + · · ·+ 1⊗ · · · ⊗ 1⊗ Fi

and similarly for EiSκi .

Standard modules

As you may have guessed

Proposition

[Sκi ] = Fiκ(1)−1 · · ·Fi1v1 ⊗ · · · ⊗ Fin · · ·Fκ(`)vn

This makes standard modules invaluable as “test objects” for functors to seethat they behave correctly on the Grothendieck group.

For example, FiSκi has a filtration which categorifies the usual formula

∆(`)(Fi) = Fi ⊗ K̃i ⊗ · · · ⊗ K̃i + · · ·+ 1⊗ · · · ⊗ 1⊗ Fi

and similarly for EiSκi .

Braiding functors Definition

Derived category

What functors? Well, we had a whole lot of maps earlier, corresponding toany tangle (though it was enough to define them for small pictures).

Unfortunately, if we want to categorify these using the yoga we’ve used thusfar, we run into a problem: the coefficients aren’t positive.

If you want to have a “direct minus” in a category, you have to use some kindof category of complexes. We let Vλ be the bounded-above derived categoryof Tλ -mod.

I bet lots of you are happier with the homotopy category, but that doesn’twork so well for me. Working in that category would require me knowingsome projective resolutions that are very hard to write down.

Derived category

What functors? Well, we had a whole lot of maps earlier, corresponding toany tangle (though it was enough to define them for small pictures).

Unfortunately, if we want to categorify these using the yoga we’ve used thusfar, we run into a problem: the coefficients aren’t positive.

If you want to have a “direct minus” in a category, you have to use some kindof category of complexes. We let Vλ be the bounded-above derived categoryof Tλ -mod.

I bet lots of you are happier with the homotopy category, but that doesn’twork so well for me. Working in that category would require me knowingsome projective resolutions that are very hard to write down.

Braiding and duals

TheoremGiven any sequence λ:

For any `-strand braid σ, we have a functor Vλ → Vσλ which inducesthe usual braided structure on the GG.

For any λ, and λ+ given by adding an adjacent pair of dual highestweights, we have functors Vλ+ → Vλ inducing evaluation and quantumtrace on GG, and dually for coevaluation and quantum cotrace (but for afunny ribbon structure!).

My goal for the rest of this talk is to describe these functors, and how theygive knot invariants.

Braiding and duals

Braiding

So, now we need to look for braiding functors.

Consider the bimodule Bi over Tλ and T(i,i+1)·λ given by exactly the samesort of diagrams, but with a single crossing inserted between the ith andi + 1st crossings.

Theorem

The derived tensor product −L⊗Tλ Bi : Vλ → V(i,i+1)·λ categorifies the

braiding map Ri : Vλ → V(i,i+1)·λ. The inverse functor is given byRHom(Bi,−).

Braiding

Theorem

Braiding

Theorem

Braiding

So, firstly, what does derived tensor product mean? It means, amongst otherthings, that we could take a projective resolution of Bi as a bimodule. Thiswill be a complex in the category Tλ ⊗ Tλ -pmod which is unique up tohomotopy.

Unfortunately, I don’t understand at the moment how to write down thiscomplex explicitly. In most cases, it must have infinite length and is quitecomplex, but it would facilitate computation quite a bit.

On the other hand, part of the magic of homological algebra is that you canfigure some things out without knowing this.

Braiding functors Checking properties

Braiding

In particular, how does one check that it actually acts as the braiding? Bylooking at test objects.

Note that Vλ1 ⊗ Vλ2 is generated over Uq(g) by vectors of the form v⊗ vhigh

and under the braiding, these are sent to q?vhigh ⊗ v. As we know, thesevectors are categorified by standard modules of the form S0,n

Proposition

B1L⊗ S(0,n)i

∼= S(0,0)i (?)

Proof: · · · · · ·

· · ·

Braiding

In particular, how does one check that it actually acts as the braiding? Bylooking at test objects.

Note that Vλ1 ⊗ Vλ2 is generated over Uq(g) by vectors of the form v⊗ vhigh

and under the braiding, these are sent to q?vhigh ⊗ v. As we know, thesevectors are categorified by standard modules of the form S0,n

Proposition

B1L⊗ S(0,n)i

∼= S(0,0)i (?)

Proof: · · · · · ·

· · ·

Braiding

In particular, how does one check that it actually gives a braid groupoidaction?

The positive and negative twists are inverse because they are adjoint andderived equivalences. (Not easy! Must show that half twist sendsprojectives to tiltings.)Homological algebra song and dance: for reduced expression in thesymmetric group, its positive lift to a braid sends projectives to modules.So we just have to check that as modulesBi ⊗T Bi+1 ⊗T Bi ∼= Bi+1 ⊗T Bi ⊗T Bi+1

Braiding

Coevalution and quantum trace

We also need functors corresponding to the cups and caps in our theory.First, consider the case where we have two highest weights λ and−w0λ = λ∗. We must first define an isomorphism between Vλ∗ and V∗λ. Thatis to say, a pairing Vλ × Vλ∗ → C(q).

We start with a chosen highest weight vector of both representations vλ, vλ∗(this comes from the irrep in Tλλ -mod ∼= k -mod). So, a pairing is fixed by achoice of lowest weight vector.

Pick a reduced expression

w0 = s1 · · · sn with corresponding roots α1, · · · , αn.

Then we have a lowest weight vector of the form

vlow = F(α∨n (sn−1···s1λ))in · · ·F(α∨2 (s1λ))

i2 F(α∨1 (λ))i1 vλ

We will always choose this one.Ben Webster (UVA) HRT : Lecture VIII April 8, 2014 26 / 36

We also need functors corresponding to the cups and caps in our theory.First, consider the case where we have two highest weights λ and−w0λ = λ∗. We must first define an isomorphism between Vλ∗ and V∗λ. Thatis to say, a pairing Vλ × Vλ∗ → C(q).

We start with a chosen highest weight vector of both representations vλ, vλ∗(this comes from the irrep in Tλλ -mod ∼= k -mod). So, a pairing is fixed by achoice of lowest weight vector.

Pick a reduced expression

w0 = s1 · · · sn with corresponding roots α1, · · · , αn.

Then we have a lowest weight vector of the form

vlow = F(α∨n (sn−1···s1λ))in · · ·F(α∨2 (s1λ))

i2 F(α∨1 (λ))i1 vλ

We will always choose this one.Ben Webster (UVA) HRT : Lecture VIII April 8, 2014 26 / 36

Invariants

We should look for a categorification of the unique invariant vectorc ∈ Vλ ⊗ Vλ∗ . We can actually guess quite easily what this should be.

The space of invariants is orthogonal under the Euler form to all projectives ofthe form FiM for any i. We know by counting arguments that all but oneindecomposable projective is a summand of a FiM.

We actually know exactly what this remaining projective Pλ is; it correspondsto the sequence of weights and roots

(λ, α(α∨1 (λ))1 , α

(α∨2 (s1λ))2 , . . . , α

(α∨n (sn−1···s1λ))n , λ∗).

So, an element of invariants is given by the simple quotient of Pλ. Denote thisLλ.

It’s pretty easy to check by hand that Lλ is killed by all Ei.

Invariants

We should look for a categorification of the unique invariant vectorc ∈ Vλ ⊗ Vλ∗ . We can actually guess quite easily what this should be.

The space of invariants is orthogonal under the Euler form to all projectives ofthe form FiM for any i. We know by counting arguments that all but oneindecomposable projective is a summand of a FiM.

We actually know exactly what this remaining projective Pλ is; it correspondsto the sequence of weights and roots

(λ, α(α∨1 (λ))1 , α

(α∨2 (s1λ))2 , . . . , α

(α∨n (sn−1···s1λ))n , λ∗).

So, an element of invariants is given by the simple quotient of Pλ. Denote thisLλ.

It’s pretty easy to check by hand that Lλ is killed by all Ei.

Coevalution and evaluation

The coevaluation functor is categorified by the functorV∅ ∼= Vect→ Vλ,λ∗ sending C→ Lλ.

The evaluation functor is categorified by

RHom(Lλ,−)[2ρ∨(λ)](2〈λ, ρ〉) : Vλ,λ∗ → V∅ ∼= Dfd(Vect).

Now, we know that if we want quantum trace, we should compromise between

Lλ[2ρ∨(λ)](2〈λ, ρ〉) and Lλ[−2ρ∨(λ)](−2〈λ, ρ〉)

DefinitionThe positive ribbon twist acts on the category by [2ρ∨(λ)](2〈λ, ρ〉).

Coevalution and evaluation

The coevaluation functor is categorified by the functorV∅ ∼= Vect→ Vλ,λ∗ sending C→ Lλ.

The evaluation functor is categorified by

RHom(Lλ,−)[2ρ∨(λ)](2〈λ, ρ〉) : Vλ,λ∗ → V∅ ∼= Dfd(Vect).

Now, we know that if we want quantum trace, we should compromise between

Lλ[2ρ∨(λ)](2〈λ, ρ〉) and Lλ[−2ρ∨(λ)](−2〈λ, ρ〉)

DefinitionThe positive ribbon twist acts on the category by [2ρ∨(λ)](2〈λ, ρ〉).

Ribbon structure

So this decategorifies to (−1)2ρ∨(λ)q2〈λ,ρ〉. Note: this is a strange ribbonelement! (It appeared in work of Snyder and Tingley on half-twist elements.)

For each ribbon element, there is a notion of “quantum dimension,” and in thispicture, qdimV|q=1 = (−1)2ρ∨(λ) dim V . For example, in sl2,

qdimVn = (−1)n qn+1 − q−n−1

q− q−1 .

From now on, all my knots are ribbon knots (in the blackboard framing), andI’ll really get invariants of ribbon knots (but twists just give grading shifts).

In particular, the algebra (which is the invariant of the circle)

Aλ = Ext•(Lλ,Lλ)[2ρ∨(λ)](2〈λ, ρ〉)

has graded Euler characteristic given by the quantum dimension of Vλ.

If Vλ is miniscule, then everything works beautifully. The dimension of Aλ isreally the dimension of Vλ. In particular, if λ = ωi for g = sln, thenAλ ∼= H∗(Grass(i, n)).

Conjecture

If λ is miniscule, Aλ ∼= H∗(Grλ).

On the other hand, if λ is not miniscule, things blow up. For example, ifg = sl2 and λ = 2, then

∑i,j(−t)j dimq Aj

λ 6= q−2t2 + 1 + q2t−2∑i,j(−t)j dimq Aj