Sweedler Notation Bialgebras and monoidal categories Rigid categories

Lecture 2

Daniel Bump

May 23, 2019

V ∗ V

K V V ∗

K

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Review of bialgebras and Hopf algebras

Recall that a bialgebra over a field K is a Vector space V withlinear maps

µ : H ⊗ H → H, ∆ : H → H ⊗ H,

η : K → H, ε : H → K ,

subject to the axioms. Associativity and coassociativity:

H ⊗ H ⊗ H H ⊗ H

H ⊗ H H

µ⊗1

1H⊗µ µ

µ

H H ⊗ H

H ⊗ H H ⊗ H ⊗ H

∆

∆ 1H⊗∆

∆⊗1H

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Bialgebra axioms (continued)

Unit:H ⊗ H

H ⊗ K H

µ

∼=

1H⊗η

H ⊗ H

K ⊗ H H

µ

∼=

η⊗KH

Counit:

H ⊗ H

H ⊗ K H

1H⊗ε∼=

∆

H ⊗ H

K ⊗ H H

ε⊗IH∼=

∆

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Bialgebra axioms (continued)

The augmentation and coaugmentation axioms:

H ⊗ H K × K

H K

ε×ε

µ ∼=

ε

K H

K × K H × H

η

∼= ∆

η×η

These say that the counit is an algebra homomorphism, andthat the unit is a coalgebra homomorphism, respectively.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Bialgebra axioms (concluded)

Let τ(a⊗ b) = b ⊗ a. Hopf:

H ⊗ H H ⊗ H ⊗ H ⊗ H H ⊗ H ⊗ H ⊗ H

H H ⊗ H

∆⊗∆

µ

1H⊗τ⊗1H

µ⊗µ

∆

The associativity and unit axioms: H is an algebra.The coassociativity and counit: H is an coalgebra.Augmentation axiom: counit is an algebra homomorphismDually, the unit is a coalgebra homomorphismThe Hopf axiom: ∆ is an algebra homomorphismEquivalently µ is a coalgebra homomorphism

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Definition of a Hopf algebra

A Hopf algebra is a bialgebra with a linear map S : H → Hcalled the antipode satisfying

H H ⊗ H H ⊗ H

K H

∆

ε

1H⊗S

µ

η

H H ⊗ H H ⊗ H

K H

∆

ε

S⊗1H

µ

η

In the analogy between Hopf algebras and groups, thissubstitutes for g · g−1 = g−1 · g = 1.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Sweedler Notation (I)

We will use ordinary ring notation for the multiplication and unitµ and η. Thus if a,b ∈ H let a · b = µ(a⊗ b) and let 1H = η(1K ).Indeed we may identify K with a subring of the center of Husing η.

The Sweedler notation streamlines the formula

∆(a) =N∑

i=1

a′i ⊗ a′′i .

Instead, write∆(a) = a(1) ⊗ a(2).

We are omitting the summation from the notation.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Sweedler Notation (II)

Sweedler notation from the previous slide:

∆(a) = a(1) ⊗ a(2). (1)

The co-associativity

(∆⊗ 1H)∆(a) = (1⊗∆)∆(a)

means

a(1)(1) ⊗ a(1)(2) ⊗ a(2) = a(1) ⊗ a(2)(1) ⊗ a(2)(2).

We will writea(1) ⊗ a(2) ⊗ a(3)

for either of these decompositions. Note that a(2) has a differentmeaning here than it did in (1).

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Axioms in Sweedler notation (I)

The counit axiom:

H ⊗ H

H ⊗ I H

1H⊗ε∼=

∆

H ⊗ H

I ⊗ H H

ε⊗IH∼=

∆

In Sweedler notation,

a = a(1)ε(a(2)) = ε(a(1))a(2).

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Axioms in Sweedler notation (II)

The antipode axiom:

H H ⊗ H H ⊗ H

K H

∆

ε

1H⊗S

µ

η

H H ⊗ H H ⊗ H

K H

∆

ε

S⊗1H

µ

η

ε(a) = a(1)S(a(2)) = S(a(1))a(2).

We should writeηε(a) = · · ·

but we identify ε(a) ∈ K with its image under η.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Hopf axiom

If A and B are algebras, so is A⊗ B with multiplication

(a⊗ b)(a′ ⊗ b′) = (aa′)⊗ (bb′).

If µA, µB and µA⊗B are the multiplications in A, B and A⊗ B thismeans

µA⊗B = (µA ⊗ µB) ◦ (1A ⊗ τ ⊗ 1B).

Hence we write the Hopf axiom

H ⊗ H (H ⊗ H)⊗ (H ⊗ H)

H H ⊗ H

∆⊗∆

µH µH⊗H

∆

In other words, ∆ : H → H ⊗ H is an algebra homomorphism.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Hopf axiom in Sweedler notation

We may also translate the Hopf axiom into Sweedler notation.

H ⊗ H (H ⊗ H)⊗ (H ⊗ H)

H H ⊗ H

∆⊗∆

µH µH⊗H

∆

Applying this to x ⊗ y ∈ H ⊗ H gives

(xy)(1) ⊗ (xy)(2) = x(1)y(1) ⊗ x(2)y(2).

Sweedler Notation Bialgebras and monoidal categories Rigid categories

The antipode

PropositionIn a Hopf algebra S(ab) = S(b)S(a).

The proof of this will be good practice for Sweedler notation sowe will explain it carefully. This is an analog of the groupproperty (xy)−1 = y−1x−1 so the problem will be to translatethe proof of that fact into the Hopf algebra setting. Let usponder the middle expression in

(xy)−1 = (xy)−1xyy−1x−1 = y−1x−1.

The Hopf analog is

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3))

where

x(1) ⊗ x(2) ⊗ x(3) = (∆⊗ 1)∆(x) = (1⊗∆)∆(x)

Sweedler Notation Bialgebras and monoidal categories Rigid categories

The antipode (continued)

The Hopf analog of (xy)−1xyy−1x−1 is

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) (2)

where

x(1) ⊗ x(2) ⊗ x(3) = (∆⊗ 1)∆(x) = (1⊗∆)∆(x).

So imitating what we did for the group identity, we will unravelthis expression two ways, proving both

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(xy)

andS(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(y)S(x).

Using the antipode axiom y(1)S(y(2)) = ε(y)

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(x(1)y(1))x(2)ε(y(2))S(x(3)).

What just happened?

Sweedler Notation Bialgebras and monoidal categories Rigid categories

What just happened?

Here we are parsing

y(1) ⊗ y(2) ⊗ y(3) = (1⊗∆)∆(y) = y(1) ⊗ y(2)(1) ⊗ y(2)(2).

Now applying the map a⊗ b ⊗ c 7→ a⊗ bS(c) to both sides ofthis identity gives

y(1) ⊗ y(2)S(y(3)) = y(1) ⊗ S(y(2)(1))y(2)(2) = y(1) ⊗ ε(y(2)).

That is how we get

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(x(1)y(1))x(2)ε(y(2))S(x(3)).

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Going back to (2), bear in mind that ε(a) is a scalar and can bemoved around at will in formulas. Also ε(a)ε(b) = ε(ab) sincethe counit is a ring homomorphism by the augmentation axiom.

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) =S(x(1)y(1))x(2)ε(y(2))S(x(3)) =S(x(1)y(1))x(2)S(x(3))ε(y(2)) =S(x(1)y(1))ε(x(2))ε(y(2)) =S(x(1)y(1)ε(x(2)y(2))).

Now by the Hopf axiom

x(1)y(1) ⊗ x(2)y(2) = (xy)(1) ⊗ (xy)(2)

so

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S((xy)(1)ε((xy)(2))) = S(xy).

Sweedler Notation Bialgebras and monoidal categories Rigid categories

The other side

Now let us unravel (1) a different way.

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) =S(x(1)(1)y(1)(1))x(1)(2)y(2)(2)S(y(2))S(x(2)) = Hopf axiomS((x(1)y(1))(1))(x(1)y(1))(2)S(y(2))S(x(2)) =ε(x(1)y(1))S(y(2))S(x(2)) =S(ε(y(1))y(2))S(ε(x(1))x(2))

and so

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(y)S(x).

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Slogans

Modules over a bialgebra are a monoidal categoryFinite-Dimensional Modules over a Hopf algebra are a rigidmonoidal categoryModules over a quasitriangular bialgebra are a braidedmonoidal categoryFinite-Dimensional Modules over a ribbon Hopf algebra area ribbon category

For the first item, if V and W are modules over H, then V ⊗Wis a module over H ⊗ H. Since ∆ is an H-modulehomomorphism H → H ⊗ H, we may use it to transport thismodule structure to H. In Sweedler notation

a(v ⊗ w) = a(1)v ⊗ a(2)w .

The coassociativity means that U ⊗ (V ⊗W ) and (U ⊗ V )⊗Whave the same module structure.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Bialgebras and monoidal categories

We have just seen that the category of modules over abialgebra is a monoidal category, and that the key to this is thecomultiplication.

There is a dual construction. Since the axioms of a bialgebraare self-dual, this statement has a dual one, that the categoryof comodules over a bialgebra is a monoidal category.

This fact arises in the theory of affine algebraic groups. If G isan affine group scheme over a field K , the affine algebra O(G)is a commutative Hopf algebra, and if V is a module for G, thenthe dual space of V is a comodule for O(G). For a proof, seeWaterhouse, Introduction to Affine Group Schemes Section 3.2.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Dual objects

Our second slogan is that modules over a Hopf algebra form arigid category. This is a category in which objects have duals.The archetype is the category finite-dimensional vector spacesover a field K .

Let V be an object in a monoidal category with unit object I. Wewill abstract the properties of a (left) dual. This comes withmorphisms evV : V ∗ ⊗ V → I and coevV : I → V ⊗ V ∗ calledevaluation and coevaluation.

The evaluation and coevaluation maps are subject to thefollowing axioms.

(1V⊗evV )◦(coevV ⊗1V ) = 1V , (evV ⊗1V∗)◦(1V∗⊗coevV ) = 1V∗ .

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Example: finite-dimensional vector spaces

In the category of finite-dimensional vector spaces, I = KV ∗ = Hom(V ,K ) and evV (v∗ ⊗ v) = v∗(v), evaluating thelinear functional v∗ ∈ V ∗ at the vector v .

Continuing with the example of a finite-dimensional vectorspace, we have to define the coevaluation, which is now to be alinear map K → V ⊗ V ∗. We pick dual bases vi of V and v∗i ofV ∗ and define coevV (a) = a

∑vi ⊗ v∗i . You may check that this

is independent of V , and that the axioms are satisfied.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Diagrams for evaluation and coevaluation

We can represent evV : V ∗ ⊗ V → K and coevV : K → V ∗ ⊗ VThe diagram is to be read from top to bottom.

V ∗ V

K V V ∗

K

At the top of the diagram is a sequence of spaces which are tobe tensored together, hence V ∗ ⊗V in the left figure; so the twofigures represent morphisms V ∗ ⊗ V → K and K → V ⊗ V ∗

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Diagrams for rigidity axioms (I)

The first axiom

(1V ⊗ evV ) ◦ (coevV ⊗1V ) = 1V

means that the following two diagram represents the identitymap V → V .

V K

K V

V V ∗ V

Here coevV ⊗1V : V = K ⊗ V → V ⊗ V ∗ ⊗ V and1V ⊗ evV : V ⊗ V ∗ ⊗ V → V ⊗ K = V .

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Diagrams for rigidity axioms (II)

Identifying K ⊗ V = V ⊗ K in the above diagram, the K issuperfluous. Moreover we may rely on the reader to slice thediagram horizontally at intervals to obtain the sequence ofspaces V , V ⊗ V ∗ ⊗ V ; we have only to label the segments ofthe diagram.

V

V

V ∗

V

V

Then the axiom asserts that the above two morphisms areequal. (The second one is the identity map V → V .)

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Diagrams for rigidity axioms (III)

Here is the dual axiom

(evV ⊗1V∗) ◦ (1V∗ ⊗ coevV ) = 1V∗ .

V ∗

V ∗

V

V ∗

V ∗

We see that the two axioms may be understood as assertingthat such bends may be straightened.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Uniqueness of the left dual

Suppose we have another left dual V ∗. Let evV and coevV bethe corresponding evaluation and coevaluation maps as Wecan construct morphisms φ : V ∗ → V ∗ and ψ : V ∗ → V ∗ by

φ = (evV ⊗IV∗)(1V∗ ⊗ coevV )

ψ = (evV ⊗ IV∗)(1V∗ ⊗ coevV )

V ∗

V ∗

V

V ∗

V ∗

V

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Uniqueness of the dual (continued)

We may diagram the composition ψ ◦ φ:

V ∗

V

V

V ∗

V ∗

= V ∗

V

V

V ∗V ∗

V ∗

ψ ◦ φ = (evV ⊗IV∗)(1V∗ ⊗ coevV )(evV ⊗ IV∗)(1V∗ ⊗ coevV )

= (evV ⊗IV∗)(IV∗⊗IV⊗evV⊗IV∗)(IV∗⊗coevV⊗IV⊗IV∗)(IV∗⊗coevV )

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Uniqueness of the dual (continued)

After switching the order of coevV and evV as above (just usingthe fact that ⊗ is a bifunctor) we may use the axioms twice toconclude that ψ ◦ φ = IV∗ :

V ∗

V

V ∗

=

V ∗

V ∗

Similarly φ ◦ ψ = IV∗ , so φ and ψ are inverse isomorphisms.This proves the uniqueness of the dual, and also illustrates theuse of diagrammatic methods in proofs.

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Dual morphisms and rigid categories

We define an object V in a monoidal category to be rigid if ithas a left dual V ∗. The category is called rigid if every object is.

Suppose that V and W are objects in a rigid category andf : V →W is a morphism. Define a morphism f ∗ : W ∗ → V ∗ by

f ∗ = (evW ⊗1V∗)(1W∗ ⊗ f ⊗ 1V∗)(IW∗ ⊗ coevV ).

Representing f as a dot (left) then f ∗ is as on the right.

V

W V ∗

W ∗

W

V

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Exercises

Exercise 1. Let f : V →W and g : W → U be morphisms in arigid category. Prove that (gf )∗ = f ∗g∗.

Let V be a vector space and V ∗ its dual. We will write 〈v∗, v〉instead of v∗(v) for the dual pairing.

Exercise 2. Let H be a Hopf algebra, V a finite-dimensionalmodule, and let V ∗ be its dual vector space. If a ∈ H andv∗ ∈ V ∗ define av∗ by

〈av∗, v〉 = 〈v∗,S(a)v〉.

Prove that this makes V ∗ into a module over H.

Exercise 3. Prove that the category of finite-dimensionalmodules over a Hopf algebra is a rigid category.