Lecture 2 - Stanford Universitysporadic.stanford.edu/quantum/lecture2.pdf · Lecture 2 Daniel Bump...
Transcript of Lecture 2 - Stanford Universitysporadic.stanford.edu/quantum/lecture2.pdf · Lecture 2 Daniel Bump...
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Sweedler Notation Bialgebras and monoidal categories Rigid categories
Lecture 2
Daniel Bump
May 23, 2019
V ∗ V
K V V ∗
K
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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
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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∼=
∆
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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.
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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
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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.
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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.
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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).
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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).
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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.
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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.
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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).
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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)
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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?
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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)).
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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).
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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).
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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.
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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.
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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∗ .
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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.
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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 ∗
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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 .
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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 .)
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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.
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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
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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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Sweedler Notation Bialgebras and monoidal categories Rigid categories
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.
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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
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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.