Lecture 3 - Stanford Universitysporadic.stanford.edu/quantum/lecture3.pdf · Lecture 3 Daniel Bump May 24, 2019 W VWV W VWV. Braided categoriesQuasitriangular Hopf algebrasQuantized

Braided categories Quasitriangular Hopf algebras Quantized enveloping algebras Reidemeister moves Braiding in a rigid category

Lecture 3

Daniel Bump

May 24, 2019

W

V ∗ W V

W

V ∗ W V


Monoidal categories

The axioms for a braided monoidal category are due to Joyaland Street. Braided Monoidal Category (Wikipedia Link)

Let C be a monoidal category. We recall that if A,B,C areobjects in C then there are natural isomorphisms(A⊗ B)⊗ C ∼= A⊗ (B ⊗ C). We will not distinguish betweenthese objects and just denote either as A⊗ B ⊗ C. It is aconsequence of Mac Lane’s coherence theorem that thisidentification will never lead to any difficulties. We will neverworry about this point.

https://en.wikipedia.org/wiki/Braided_monoidal_category


Braided categories

In a braided category there are explicit braid isomorphismscA,B : A⊗ B → B ⊗ A but now we must be careful. For examplethe composition cB,AcA,B is not assumed to be the identity. SocA,B and c−1

B,A are distinct isomorphisms A→ B.

We will notate the morphism cA,B by an over crossing and cB,Aby an under crossing.

A B

B A

cA,B

A B

B A

c−1B,A


Naturality

We review the important notion of a natural transformation. Weused this implicitly when we defined a monoidal category inLecture 1, where we said that the isomorphisms

(A⊗ B)⊗ C ∼= A⊗ (B ⊗ C)

are required to be natural.

This means, explicitly, the following. Since ⊗ is a bifunctor, ifα : A→ A′, β : B → B′ and γ : C → C′ are morphisms then wehave on the left and right of the following diagram.

(A⊗ B)⊗ C A⊗ (B ⊗ C)

(A′ ⊗ B′)⊗ C′ A′ ⊗ (B′ ⊗ C′)

∼=

(α⊗β)⊗γ α⊗(β⊗γ)

∼=


Naturality of the commutativity map

The first axiom of a braided category is that the morphismscA,B : A⊗ B → B ⊗ A are to be natural. This means that ifα : A→ A′ and β : B → B′ are morphisms, then

(β ⊗ α) ◦ cA,B = cA′,B′ ◦ (α⊗ β)

α β

A B

B′ A′

= β α

A B

B′ A′

(We are representing the morphisms α, β by dots.)


Braided category axioms

The braid morphism cA,B : A⊗ B → B ⊗ A is sometimes calledan R-matrix. It is assumed to satisfy:

A⊗ B ⊗ C B ⊗ C ⊗ A

B ⊗ A⊗ C

cA,B⊗C

cA,B⊗1C 1B⊗cA,C

B C A

A B C

=

B ⊗ C A

A B ⊗ C


Mirror image axiom

A⊗ B ⊗ C C ⊗ A⊗ B

A⊗ C ⊗ B

cA×B,C

1A⊗cB,C cA,C⊗1B

C A B

A B C

=

C A⊗ B

A⊗ B C

This completes the definition of a braided monoidal category.


The Yang-Baxter equation

We will show that the Yang-Baxter equation is true in a braidedmonoidal category. Remember that this means we have toshow:

C B A

A B C

=

C B A

A B C


Proof of the Yang-Baxter equation

C B A

A B C

=

cB,C

C ⊗ B A

A B ⊗ C

= cB,C

C ⊗ B A

A B ⊗ C

=

C B A

A B C

Showing that the Yang-Baxter equation is true in a braidedmonoidal category uses both axioms and naturality.


Slogan, revisited

Recall the slogan: modules over a quasitriangular bialgebra area braided category.

We wish to impose properties on a bialgebra H that will makethe module category into a braided modular category.

Thus we will arrive at the notion of a quasitriangular or braidedHopf algebra (QTHA), introduced by Drinfeld in his 1986 ICMlecture.


Quasitriangular Hopf algebras

Let R be an element of H ⊗ H, which we assume to beinvertible. If U and V are H-modules, consider the mapτR : U ⊗V → V ⊗U. (τ is always the flip map x ⊗ y → y ⊗ x ofany tensor product U ⊗ V .)

PropositionSuppose that for x ∈ H we have

R∆(x)R−1 = τ∆(x).

Then for all U,V, the map cU,V : U ⊗ V → V ⊗ U defined bycu,v (u ⊗ v) = τR(u ⊗ v) is an H-module homomorphism.


Proof

We are assuming R∆(x)R−1 = τ∆(x) which we rewrite

R∆(x) =(τ(∆(x))

)R.

We will writeR = R(1) ⊗ R(2),

where we have omitted an implicit summation. Combining thiswith ∆(x) = x(1) ⊗ x(2) (Sweedler notation)

R(1)x(1) ⊗ R(2)x(2) = x(2)R(1) ⊗ x(1)R(2)

in other words

R(1)x(1) ⊗ R(2)x(2) = x(2)R(1) ⊗ x(1)R(2). (1)


Proof (continued)

Now suppose u ⊗ v ∈ U ⊗ V . We need to show for x ∈ H:

τR(x(u ⊗ v)) = xτR(u ⊗ v)

in other words

τ(R(1)x(1)u ⊗ R(2)x(2)v) = (x(1) ⊗ x(2))τ(R(1)u ⊗ R(2)v)

orR(2)x(2)v ⊗ R(1)x(1)u = x(1)R(2)v ⊗ x(2)R(1)u

This follows from (1) by applying the map

a⊗ b 7→ bv ⊗ au

and we are done.


Some notation

Remember that R ∈ H ⊗ H. We have written

R = R(1) ⊗ R(2),

and this is shorthand for a sum, say

R =N∑

i=1

(R′i )⊗ (R′′i ),

where we are writting R(1) instead of R′i and R(2) instead of R′′i .We will consider some elements of H ⊗ H ⊗ H,

R12 = R(1) ⊗ R(2) ⊗ 1R, R13 = R(1) ⊗ 1R ⊗ R(2),

R23 = 1R ⊗ R(1) ⊗ R(2).


Quasitriangular Hopf algebras

Definition (Drinfeld)A quasitriangular or braided Hopf algebra is a Hopf algebra Hwith R ∈ H ⊗ H such that

R∆(x)R−1 = τ∆(x)

and(∆⊗ 1)R = R13R23, (1⊗∆)R = R13R12.

The element R is called the universal R-matrix. We understandthe significance of the first condition: it means that if U,V areH-modules, then cU,V : U ⊗ V → V ⊗ U defined by(u, v) 7→ τR(u ⊗ v) is an H-module homomorphism. Whatabout the other properties?


Braided category axioms and quasitriangularity

We will show that the axiom

(1⊗∆)R = R13R12

is equivalent to the axiom

A⊗ B ⊗ C B ⊗ C ⊗ A

B ⊗ A⊗ C

cA,B⊗C

cA,B⊗1C 1B⊗cA,C

First let us argue that the top arrow cA,B⊗C is

cA,B⊗C(a,b, c) = θ((1⊗∆)R)(a⊗ b ⊗ c),

whereθ(a⊗ b ⊗ c) = b ⊗ c ⊗ a.


Braided category axiom continued

From the previous slide, we are checking:

cA,B⊗C(a,b, c) = θ((1⊗∆)R)(a⊗ b ⊗ c), (2)

whereθ(a⊗ b ⊗ c) = b ⊗ c ⊗ a.

Indeed let us treat d = b ⊗ c as a unit and remember thedefinition of cA,D, with D = B ⊗ C.

cA,B⊗C(a⊗ d) = τR(a⊗ d) = τ(R(1)a⊗ R(2)d , )

and θ is just the map τ in this setting. Now since themultiplication of Hopf elements on a tensor product of modulesis through the tensor product,

R(2)d = ∆(R(2))(b ⊗ c).

Thus we obtain (2).


Braided category axiom continued

Now we want to prove:

A⊗ B ⊗ C B ⊗ C ⊗ A

B ⊗ A⊗ C

θ((1⊗∆)(R))

(τ⊗1C)R12 (1B⊗τ)R23

We need to show that

θ((1⊗∆)R) = (1B ⊗ τ)R12(τ ⊗ 1C)R12.

We have(τ ⊗ 1C)R23(τ ⊗ 1C) = R13

so because (1B ⊗ τ)(τ ⊗ 1C) = θ:

(1B ⊗ τ)R23(τ ⊗ 1C)R12 = (1B ⊗ τ)(τ ⊗ 1C)R13R23 = θR13R23


Braided category axiom, concluded

Now using the assumption from the definition of a QTHA:

(1⊗∆)R = R13R23,

the commutativity of

A⊗ B ⊗ C B ⊗ C ⊗ A

B ⊗ A⊗ C

θ((1⊗∆)(R))

(τ⊗1C)R12 (1B⊗τ)R23

follows and we have proved one of the braided categoryaxioms.That the mirror image axiom and naturality are satisfied are leftto the exercises.


Example of a QTHA

Let G be the finite cyclic group of order n:

〈g|gn = 1〉

Let H = C[G] be the group algebra and let q = e2πi/n. It is aHopf algebra with comultiplication

∆(ga) = ga ⊗ ga.

DefineR =

1n

∑a,b

q−abga ⊗ gb ∈ H ⊗ H.

(Sum is over a,b mod n). This is invertible with


R is invertible

R =1n

∑a,b

q−abga ⊗ gb ∈ H ⊗ H.

R−1 =1n

∑a,b

qabga ⊗ gb.

To see this, the product of these two elements is

1n2

∑a,b,c,d

q−ab+cdga+c⊗gb+d =1n2

∑t ,u

∑a,b

q−ab+(t−a)(u−b)

gt⊗gu.

The inner sum is

qtu∑a,b

q−tb−ua =

{n2 if t = u = 0,0 otherwise.

From this it follows that R−1 is indeed an inverse to R.


Cyclic group QTHA

TheoremH is quasitriangular with universal R-matrix R.

Proof: The axiom R∆(h)R−1 = τ∆(h) for h ∈ H is trival sinceH is both commutative and cocommutative. We have

R13R12 =1n2

∑a,b

q−abga ⊗ 1⊗ gb

∑c,d

q−cdgc ⊗ gd ⊗ 1

=

1n2

∑a,b,c,d

q−ab−cdga+c ⊗ gd ⊗ gb


Proof, concluded

This equals

1n2

∑t ,b,d

(∑a

q−ab−(t−a)d

)gt ⊗ gd ⊗ gb.

The inner sum is zero unless b = d , so this equals

1n

∑t ,b

(∑a

q−tb

)gt ⊗ gb ⊗ gb = (1⊗∆)R.

This proves R13R12 = (1⊗∆)R and R13R23 = (∆⊗ 1)R issimilar.


Quantized enveloping algebras

We have mentioned in Lecture 1 that there are two classes ofHopf algebras that may be derived from a Lie group: theenveloping algebra, and the affine algebra or coordinate ring.Both admit quantum deformations.

A particular subtle point is the existence of the universalR-matrix. If g is a complex semisimple Lie algebra andH = Uq(g) is the quantized enveloping algebra, the R-matrixdoes not live in H itself but in a completion. (So strictlyspeaking, H does not satisfy the definition of a QTHA.)

Morally, however Uq(g) is a QTHA. Moreover if q is a root ofunity Uq(g) has a finite-dimensional quotient that is strictlyquasitriangular.


Uq(sl2)

Recall that the Lie algebra sl2 is 3-dimensional with basis

E =

(0 10 0

), F =

(0 01 0

), H =

(1 00 −1

)with bracket operations

[E ,F ] = H, [H,E ] = 2E , [H,F ] = −2F .

To construct the quantized enveloping algebra we replace H byan invertible element K that you can think of as

K =

(q

q−1

)so that

KEK−1 = q2E , KFK−1 = q−2F EF − FE =K − K−1

q − q−1 .


The comultiplication

So we let Uq(sl2) be the algebra with generators K ,K−1,E andF subject to the above relations. It is not possible to naively setq = 1 and recover U(sl2) but a more careful procedure willproduce this result. This is done in Kassel’s book.

There is an algebra homomorphism ∆ : H → H ⊗ H defined by∆(K ) = K ⊗ K ,

∆(E) = E ⊗ K + 1⊗ E , ∆(F ) = F ⊗ 1 + K−1 ⊗ F ,

Indeed, it is not hard to check that the expressions on the rightsatisfy the same relations as E ,F ,K . There is a counitε : K → C defined by ε(K ) = 1, ε(E) = ε(F ) = 0 and antipodeS : K → K such that S(K ) = K−1, S(E) = −E and S(F ) = −F .So H is a Hopf algebra.


At a root of unity

We refer to Majid and Kassel for the following facts.

Let q = e2πi/n where we assume that n is odd. Then En, F n

and K n are central, and quotienting by them produces thefinite-dimensional Hopf algebra uq(sl2). It is quasitriangular withuniversal R-matrix

R =1n

n−1∑a,b=0

q−2abK a ⊗ K b

n−1∑r=0

(q − q−1)r

[r ]q−2!E r ⊗ F r

where

r ]q−2! =r∏

t=1

[t ]q−2 , [r ]q−2 =1− q−2r

1− q−2 .


Knots and links

A knot is a smooth simple closed curve in S3. A link is a finiteunion of disjoint smooth simple closed curves.

Knot Theory (Wikipedia)

Knotinfo web page

The Knot Atlas

https://en.wikipedia.org/wiki/Knot_theory

http://www.indiana.edu/~knotinfo/

http://katlas.math.toronto.edu/wiki/Main_Page


Wild knots

We only consider knots and links that are equivalent by ambientisotopy to a smooth curve or (equivalently) a finite union ofsegments, to avoid wild knots like this one:

Wild knots (Wikipedia Link)

https://en.wikipedia.org/wiki/Wild_knot


Reidemeister moves

Knots are typically studied by projecting them onto the plane. Aknot in 3-space is projected onto R2 with the crossings markedto show which strand is over and which is under.

An issue then is to determine when two knots are equivalent byan ambient isotopy (i.e. isotopy of the ambient S3).


Reidemeister moves of Type I

If the knots are represented by their two-dimensionalprojections, a necessary and sufficient condition is that theseprojections be related by a sequence of Reidemeister moves.There are three kinds. A Reidemeister move of Type I undoes atwist:


Reidemeister moves of Type II

A Reidemeister move of type II changes:


Reidemeister moves of Type III

A Reidemeister move of type III changes:


Difficulty with Type I

We note that Reidemeister moves of Types II and III haveanalogs in braided rigid categories. However a Reidemeistermove of Type I does not. In a braided rigid category, the twistthat we drew is a morphism V ∗∗ → V :

V

V ∗∗

V ∗

Since this morphism doesn’t map to the same object, we do notexpect to be able to straighten it.


Framed knots and links

Instead of working with knots, we wish to work with framed links(“ribbons”) which can also be projected to the plane, and whichare then equivalent when they are related by Reidemeistermoves of Types II and III, but not of Type I.

A framed knot is not just a strand, but a strand with a givennormal vector field. If we fatten it up in the direction of thenormal vector field, it becomes a ribbon, which can twist inspace.


Twisting

If we double the strand in order to visualize how it will behave in3 dimensions, the red strand passes first over, then under theblue strand. When we pull it straight, it gets a double twist.


Evaluation and coevaluation of tensor products

We proved in Lecture 2 that the evaluation and coevaluationcharacterize the dual uniquely.

V

V

V ∗

V

V

V ∗

V ∗

V

V ∗

V ∗

We can use this to identify evV when V = U ⊗W is a tensorproduct.


Evaluation and coevaluation of tensor products

U W

U W

U∗

W ∗ =

U W

U W

Uniqueness of the dual implies

evU⊗W = evW (1W∗ ⊗ evU ⊗1W ),

coevU⊗W = (1U ⊗ coev W ⊗ 1U∗) coevU

(Other axiom can be diagrammed similarly.)


A relation between cV ,W and cV∗,W

We will prove that

(evV ⊗1W )(1V∗ ⊗ cW ,V ) = (1W ⊗ evV )(c−1W ,V∗ ⊗ 1V )

W

V ∗ W V

W

V ∗ W V

The figures are isotopic, so this is expected, but how to proveit?


Proof

W

V ∗ W V

W

V ∗ W V

What just happened?


We used the braided category axiom and naturality

W ⊗ V ∗ ⊗ V V ∗ ⊗ V ⊗W K ⊗W = W

V ∗ ⊗W ⊗ V

cW ,V∗⊗V

cW ,V∗⊗1V

evV

1∗V⊗cW ,V

K W

W V ∗ V

=

K W

W V ∗ ⊗ V

=

K W

W V ∗ ⊗ V


Exercises

Exercise 1. In the proof that the category of modules for aQTHA is a braided modular category, we left naturality and themirror image axiom to the reader. Check these.

Exercise 2. Prove that

cV∗,W = (evV ⊗1W ⊗ 1V∗)(1V∗ ⊗ c−1V ,W )(1V∗ ⊗ 1W ⊗ coevV ).

W V ∗

WV ∗

=

V ∗

V

W

V ∗ W

Documents

Lecture 3 - Stanford Universitysporadic.stanford.edu/quantum/lecture3.pdf · Lecture 3 Daniel Bump May 24, 2019 W VWV W VWV. Braided categoriesQuasitriangular Hopf algebrasQuantized