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### Transcript of Lecture 6 - Stanford The standard module of Uq(sl2) and its R-matrixThe 6-vertex model Lecture 6...

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Lecture 6

Daniel Bump

May 28, 2019

v v v v v

ṽ ′′ v ′′ v ′ v ′ v ′ v ′ v ′

σ1 σ2 σ3 σ4 σ5

τ1 τ2 τ3 τ4 τ5

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Uq(sl2)

We remind the reader of Uq(sl2). It has generators E , F and K ; K is “group-line” and has a multiplicative inverse K−1. The relations are

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

q − q−1 .

The counit and comultiplication are

∆(K ) = K ⊗ K ,

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

ε(K ) = 1, ε(E) = ε(F ) = 0.

The antipode satisfies S(K ) = K−1, S(E) = −E , S(F ) = −F .

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Representations of SL2

With a caveat, the following representation theories are the same:

Finite-dimensional representations of SL(2,R); Finite-dimensional representations of SU(2); Finite-dimensional analytic representations of SL(2,C); Finite-dimensional complex representations of their Lie algebras sl2(R), su2, sl2(C); The enveloping algebras of these Lie algebras; The quantized enveloping algebra Uq(sl2), where q is not a root of unity.

The caveat is that these categories are the same simple objects but Uq(sl2) has an interesting comultiplication and R-matrix so they are not the same as braided categories.

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Representations of SL2 (continued)

Each of the above categories is semisimple, that is, every object is a direct sum of irreducibles.

There is one unique irreducible for every dimension. The two-dimensional standard representation V generates the category: every irreducible is a submodule of

⊗k V for some k . More precisely, the k -dimensional irreducible module is the symmetric power Symk−1(V ).

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Modules of sl2

The standard module of sl2 has two basis vectors x and y ; with respect to this basis

E = (

0 1 0 0

) , F =

( 0 0 1 0

) , H =

( 1 0 0 −1

) ,

where H = [E ,F ]. For the 4-dimensional irreducible:

E =

 0 1 0 00 0 2 0 0 0 0 3 0 0 0 0

, F =  0 0 0 03 0 0 0

0 2 0 0 0 0 1 0

, H =  3 0 0 00 1 0 0

0 0 −1 0 0 0 0 3



• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Modules of Uq(sl2)

The standard module of Uq(sl2) again has basis vectors x and y ; with respect to this basis

E = (

0 1 0 0

) , F =

( 0 0 1 0

) , =

( q 0 0 q−1

) .

For the 4-dimensional irreducible:

E = ( 0 1 0 0

0 0 [2] 0 0 0 0 [3] 0 0 0 0

) , F =

( 0 0 0 0 [3] 0 0 0 0 [2] 0 0 0 0 1 0

) , K =

( q3 0 0 0 0 q 0 0 0 0 q−1 0 0 0 0 q−3

)

where the “quantum integer” [n] = (qn − q−n)/(q − q−1).

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

The standard module of Uq(sl2)

So the quantized standard module has two basis vectors x , y

E = (

0 1 0 0

) , F =

( 0 0 1 0

) , K =

( q 0 0 q−1

) ,

where recall EF − FE = (K − K−1)/(q − q−1).

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

The R-matrix

We are not ready to prove that H = Uq(sl2) is (in a suitable sense) quasitriangular. The universal R-matrix lives not in H but in a suitable completion, and its existence is a subtle matter.

Nevertheless it is not the universal R-matrix itself, but the morphisms it induces. We will study the R-matrix for the standard module V of Uq(sl2).

We will determine the endomorphisms of V ⊗ V , then pick out the ones that satisfy the Yang-Baxter equation.

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Endomorphism of V ⊗ V

Let x , y be the standard basis of V . Recall

K (x) = qx , K (y) = q−1y , E(x) = 0, E(y) = x ,

F (x) = y , F (y) = 0,

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

We consider an endomorphism τR of V ⊗ V with respect to the basis x ⊗ x , x ⊗ y , y ⊗ x and y ⊗ y . The eigenspaces of K :

eigenvalue basis q2 x ⊗ x 1 x ⊗ y , y ⊗ x

q−2 y ⊗ y

These must be invariant under τR.

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

E and F

K (x) = qx , K (y) = q−1y , E(x) = 0, E(y) = x ,

F (x) = y , F (y) = 0

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

E(x ⊗ x) = 0 E(x ⊗ y) = x ⊗ x E(y ⊗ x) = q(x ⊗ x) E(y ⊗ y) = q−1(x ⊗ y) + y ⊗ x

F (x ⊗ x) = q−1x ⊗ y + y ⊗ x F (x ⊗ y) = y ⊗ y F (y ⊗ x) = q(y ⊗ y) F (y ⊗ y) = 0

E =

 0 1 q 0 0 0 0 q−1

0 0 0 1 0 0 0 0

 , F = 

0 0 0 0 q−1 0 0 0 1 0 0 0 0 1 q 0



• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

The R-matrix

Since τR preserves the eigenspaces it has the form( ∗ ∗ ∗ ∗ ∗

) .

Since it commutes with

E =

 0 1 q 0 0 0 0 q−1

0 0 0 1 0 0 0 0

, F = 

0 0 0 0 q−1 0 0 0 1 0 0 0 0 1 q 0

 it is easy computation to see that it is a a scalar multiple of

1 1− qb b

b 1− q−1b 1

 .

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

The Yang-Baxter equation

We must now have the Yang-Baxter equation

(τR)12(τR)23(τR)12 = (τR)23(τR)12(τR)23,

or equivalently R12R13R23 = R23R13R12.

With

τR =

 1

1− qb b b 1− q−1b

1

 , this implies b = 0,q or q−1. We discard the solution b = 0.

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

The R-matrices

We arrive at two R-matrices for Uq(sl2). We multiply the b = q solution by q−1:

R =

 q−1

1 q−1 − q 1

q−1

 . Similarly b = q−1 gives the alternative R-matrix:

R̃ =

 q

1 q − q−1 1

q

 . Both satisfy

R12R13R23 = R23R13R12 and produce H-module endomorphism of V ⊗ V .

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Dual R-matrices

Why are there two R-matrices? Let us consider the case of a QTHA H with universal R-matrix R = R(1) ⊗ R(2). Then it is easy to see from the axioms that R̃ = R−121 is also an R-matrix. (This is proved in Majid Chapter 5.)

Here R21 = R(2)⊗R(1). Now consider the case of sl2. By abuse of notation we will use the same letter R to denote both the universal R-matrix and the endomorphism of V that it induces.

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Dual R-matrices, continued

So what endomorphism of V ⊗ V does R21 induce? Note that R21 is R conjugated by the flip endomorphism τ : V → V , which respect to the basis x ⊗ x , x ⊗ y , y ⊗ x and y ⊗ y is the matrix

τ =

 1

0 1 1 0

1

 . Consistent with this, for

R =

 q−1 1 q−1 − q 1

q−1

, R̃ =  q 1 q − q−1

1 q

, we find that R̃ = τR−1τ .

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

A parametrized Yang-Baxter equation

For later reference, we make note of the following parametrized Yang-Baxter equation. Define

R(z) = xR − yR̃.

Then

R12(x , y)R13(x , z)R23(y , z) = R23(y , z)R13(x , z)R12(x , y).

If g is a simple Lie algebra there is always a parametrized Yang-Baxter equation associated with Uq(g) or more precisely its affinization Uq(ĝ). Is not always linear in R and R−121 as above. It can be quadratic: see Jimbo, Introduction to the Yang-Baxter equation, Internat. J. Modern Phys. A 4 (1989), equation (5.2).

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Statistical mechanics

In statistical mechanics one is concerned with large ensembles of states. Highly energetic states are less probable.

If T is the temperature, the probability of a state s is proportional to the Boltzmann weight e−kβ(s)/T . Here β is the energy of the state. The actual probability is

1 Z (S)

e−kβ(s)/T ,

where Z (S) =

∑ s

e−kβ(s)/T

is the partition function.

• The standard module of Uq(sl2) and its R-matrix The 6-vertex model

Solvable lattice models

Certain systems may be solved exactly. The first such example was the 2-dimensional Ising model (Onsager 1944). The six-vertex model was solved by Lieb and Sutherland in the 1960’s, then solved another way by Baxter whose method extended to the eight-verte