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

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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