psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Maximally extended sl(2|2)

as a Quantum Double. arXiv:1602.04988, Beisert, de Leeuw.

as a contraction limit. arXiv:1703.XXXX, Beisert, Hoare.

Reimar Hecht, ETH Zurich

TCD, 03.03.2017

1 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Centrally extended sl(2|2)

2 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Centrally extended sl(2|2)

Centrally extended sl(2|2) (psl(2|2)n C3) appears inHubbard Model

AdS/CFT

Special Lie algebra

psl(2|2) only simple Lie superalgebra with threefold centralextension

related to d(2, 1;↵), ↵ ! 0q-deformation Uq(psl(2|2)n C3)

Goal: Universal R-matrix.R

12

R13

R23

= R23

R13

R12

.

3 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Centrally extended sl(2|2)

Centrally extended sl(2|2) (psl(2|2)n C3) appears inHubbard Model

AdS/CFT

Special Lie algebra

psl(2|2) only simple Lie superalgebra with threefold centralextension

related to d(2, 1;↵), ↵ ! 0q-deformation Uq(psl(2|2)n C3)

Goal: Universal R-matrix.R

12

R13

R23

= R23

R13

R12

.

3 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Centrally extended sl(2|2)

Centrally extended sl(2|2) (psl(2|2)n C3) appears inHubbard Model

AdS/CFT

Special Lie algebra

psl(2|2) only simple Lie superalgebra with threefold centralextension

related to d(2, 1;↵), ↵ ! 0q-deformation Uq(psl(2|2)n C3)

Goal: Universal R-matrix.R

12

R13

R23

= R23

R13

R12

.

3 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Uq(psl(2|2)n C3)

Chevalley-Serre generators q = e~,

[Hi ,Ej ] = aijEj ,[Hi ,Fj ] = �aijFj ,

[Ei ,Fj ] = �ijqHi � q�Hiq � q�1 ,

E2

,F2

fermionic, and Cartan matrix

aij =

0

@2 �1 0�1 0 +10 +1 �2

1

A ,

Note: Cartan matrix is degenerate

C := 12

H1

+ H2

+ 12

H3

is central.

4 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Uq(psl(2|2)n C3)

Chevalley-Serre generators q = e~,

[Hi ,Ej ] = aijEj ,[Hi ,Fj ] = �aijFj ,

[Ei ,Fj ] = �ijqHi � q�Hiq � q�1 ,

E2

,F2

fermionic, and Cartan matrix

aij =

0

@2 �1 0�1 0 +10 +1 �2

1

A ,

Note: Cartan matrix is degenerate

C := 12

H1

+ H2

+ 12

H3

is central.

4 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Uq(psl(2|2)n C3)Serre relations: (Ei . Ej := EiEj � q�aijEjEi )

E1

. E3

= 0 = F3

. F1

, Ei . (Ei . E2) = 0 = Fi . (Fi . F2),

Quartic Serre relations give rise to additonal central elements P ,K

{E1

. E2

,E3

. E2

} = P , {F2

. F1

,F2

. F3

} = K .

In total three central generators

P = K = 0 ) sl(2|2); C = 0 ) psl(2|2).

Coproduct

�Hi = Hi ⌦ 1 + 1⌦ Hi , �C = C ⌦ 1 + 1⌦ C�Ei = Ei ⌦ 1 + q�Hi ⌦ Ei , �Fi = Fi ⌦ qHi + 1⌦ Fi ,�P = P ⌦ 1 + q�2C ⌦ P , �K = K ⌦ q2C + 1⌦ K

5 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Uq(psl(2|2)n C3)Serre relations: (Ei . Ej := EiEj � q�aijEjEi )

E1

. E3

= 0 = F3

. F1

, Ei . (Ei . E2) = 0 = Fi . (Fi . F2),

Quartic Serre relations give rise to additonal central elements P ,K

{E1

. E2

,E3

. E2

} = P , {F2

. F1

,F2

. F3

} = K .

In total three central generators

P = K = 0 ) sl(2|2); C = 0 ) psl(2|2).

Coproduct

�Hi = Hi ⌦ 1 + 1⌦ Hi , �C = C ⌦ 1 + 1⌦ C�Ei = Ei ⌦ 1 + q�Hi ⌦ Ei , �Fi = Fi ⌦ qHi + 1⌦ Fi ,�P = P ⌦ 1 + q�2C ⌦ P , �K = K ⌦ q2C + 1⌦ K

5 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Uq(psl(2|2)n C3)Serre relations: (Ei . Ej := EiEj � q�aijEjEi )

E1

. E3

= 0 = F3

. F1

, Ei . (Ei . E2) = 0 = Fi . (Fi . F2),

Quartic Serre relations give rise to additonal central elements P ,K

{E1

. E2

,E3

. E2

} = P , {F2

. F1

,F2

. F3

} = K .

In total three central generators

P = K = 0 ) sl(2|2); C = 0 ) psl(2|2).

Coproduct

�Hi = Hi ⌦ 1 + 1⌦ Hi , �C = C ⌦ 1 + 1⌦ C�Ei = Ei ⌦ 1 + q�Hi ⌦ Ei , �Fi = Fi ⌦ qHi + 1⌦ Fi ,�P = P ⌦ 1 + q�2C ⌦ P , �K = K ⌦ q2C + 1⌦ K

5 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Quantum Double

6 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Universal R-matrix an invertible element R 2 H⌦H, s.t�op(a)R = R�(a), 8a 2 H,

where�op := ⌧ ��, ⌧(a⌦ b) = b ⌦ a.

Furthermore

(�⌦ id)R = R13

R23

, (id⌦�)R = R13

R12

.

Imply the Yang-Baxter equation

R12

R13

R23

= R23

R13

R12

.

Given a representation ⇢ : H ! End(V)R⇢ = (⇢⌦ ⇢)R

Dual Hopf algebra H⇤non-degenerate bilinear pairing h·, ·i : H⇤ ⌦H ! C,

hfg , ai = hf , a(1)

ihg , a(2)

i, hf , abi = hf(1)

, aihf(2)

, biproduct , coproduct

7 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Universal R-matrix an invertible element R 2 H⌦H, s.t�op(a)R = R�(a), 8a 2 H,

where�op := ⌧ ��, ⌧(a⌦ b) = b ⌦ a.

Furthermore

(�⌦ id)R = R13

R23

, (id⌦�)R = R13

R12

.

Imply the Yang-Baxter equation

R12

R13

R23

= R23

R13

R12

.

Given a representation ⇢ : H ! End(V)R⇢ = (⇢⌦ ⇢)R

Dual Hopf algebra H⇤non-degenerate bilinear pairing h·, ·i : H⇤ ⌦H ! C,

hfg , ai = hf , a(1)

ihg , a(2)

i, hf , abi = hf(1)

, aihf(2)

, biproduct , coproduct

7 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Universal R-matrix an invertible element R 2 H⌦H, s.t�op(a)R = R�(a), 8a 2 H,

where�op := ⌧ ��, ⌧(a⌦ b) = b ⌦ a.

Furthermore

(�⌦ id)R = R13

R23

, (id⌦�)R = R13

R12

.

Imply the Yang-Baxter equation

R12

R13

R23

= R23

R13

R12

.

Given a representation ⇢ : H ! End(V)R⇢ = (⇢⌦ ⇢)R

Dual Hopf algebra H⇤non-degenerate bilinear pairing h·, ·i : H⇤ ⌦H ! C,

hfg , ai = hf , a(1)

ihg , a(2)

i, hf , abi = hf(1)

, aihf(2)

, biproduct , coproduct

7 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

[Drinfeld ’86]

Quantum Double D(H) generated by H and H⇤op withcross-relations �a = a

(1)

⌦ a(2)

Xx(1)

f(1)

hf(2)

, x(2)

i =X

hf(1)

, x(1)

if(2)

x(2)

, x 2 H, f 2 H⇤op

Universal R-matrix

R =X

i2Iei ⌦ e⇤i 2 D(H)⌦D(H),

where {ei}i2I ⇢ H and {e⇤i }i2I ⇢ H⇤op are dual bases.

In practice

For given H find subalgebra A ⇢ H s.t. H ⇠= D(A)

8 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

[Drinfeld ’86]

Quantum Double D(H) generated by H and H⇤op withcross-relations �a = a

(1)

⌦ a(2)

Xx(1)

f(1)

hf(2)

, x(2)

i =X

hf(1)

, x(1)

if(2)

x(2)

, x 2 H, f 2 H⇤op

Universal R-matrix

R =X

i2Iei ⌦ e⇤i 2 D(H)⌦D(H),

where {ei}i2I ⇢ H and {e⇤i }i2I ⇢ H⇤op are dual bases.

In practice

For given H find subalgebra A ⇢ H s.t. H ⇠= D(A)

8 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How it works for Uq(sl(3)) [Drinfeld ’86],[Burroughs ’90]1.) positve Borel sub-algebra Uq(b+), basis {E n1

1

E n1212

E n22

Hm11

Hm22

}[Hi ,Ej ] = aijEj , �Ej = Ej ⌦ 1 + q�Hi ⌦ Ej

2.) construct the dual Uq(b+)⇤, dual basis {(E n11

E n1212

E n22

Hm11

Hm22

)⇤}[H⇤i ,E

⇤j ] = �~�ij E⇤j , �E⇤j = E⇤j ⌦ 1 + e

Pi aij H

⇤i ⌦ E⇤j

3.) built the Quantum Double, i.e. find the cross-relations

[Ei ,E⇤j ] = �ij (e

Pk akjH

⇤k � q�Hi )

4.) Identify Uq(b+)⇤op ⇠= Uq(b�)

( issue for psl(2|2)n C3

bHj ⇠1

~X

i

aijH⇤i , Fj ⇠ E

⇤i ,

and thus Uq(sl(3)) ⇠= D(Uq(b+))/h bH � Hi5.) R-matrix eXq :=

P1n=0

Xn

[n;q]! , [n; q] :=1�qn1�q

R = e(q�q�1

)E1

⌦F1

q�2 e(q�q�1)E

12

⌦F21

q�2 e(q�q�1)E

2

⌦F2

q�2 qH1

⌦(2H1

+H2

)/3 qH2⌦(2H2+H1)/3,

9 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How it works for Uq(sl(3)) [Drinfeld ’86],[Burroughs ’90]1.) positve Borel sub-algebra Uq(b+), basis {E n1

1

E n1212

E n22

Hm11

Hm22

}[Hi ,Ej ] = aijEj , �Ej = Ej ⌦ 1 + q�Hi ⌦ Ej

2.) construct the dual Uq(b+)⇤, dual basis {(E n11

E n1212

E n22

Hm11

Hm22

)⇤}[H⇤i ,E

⇤j ] = �~�ij E⇤j , �E⇤j = E⇤j ⌦ 1 + e

Pi aij H

⇤i ⌦ E⇤j

3.) built the Quantum Double, i.e. find the cross-relations

[Ei ,E⇤j ] = �ij (e

Pk akjH

⇤k � q�Hi )

4.) Identify Uq(b+)⇤op ⇠= Uq(b�)

( issue for psl(2|2)n C3

bHj ⇠1

~X

i

aijH⇤i , Fj ⇠ E

⇤i ,

and thus Uq(sl(3)) ⇠= D(Uq(b+))/h bH � Hi5.) R-matrix eXq :=

P1n=0

Xn

[n;q]! , [n; q] :=1�qn1�q

R = e(q�q�1

)E1

⌦F1

q�2 e(q�q�1)E

12

⌦F21

q�2 e(q�q�1)E

2

⌦F2

q�2 qH1

⌦(2H1

+H2

)/3 qH2⌦(2H2+H1)/3,

9 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

2.) construct the dual Uq(b+)⇤

Calculate the product of f , g 2 Uq(b+)⇤w.r.t. ei 2 {En1

1

En1212

En22

Hm11

Hm22

}

fg =X

i2Ihfg , ei ie⇤i =

X

i2Ihf , (ei )

(1)

ihg , (ei )(2)

ie⇤i ,

e.g.

H⇤i E⇤j =

E⇤j H⇤i =

Calculate the coproduct of f 2 H⇤

�(f ) =X

i ,j2Ih�(f ), ei ⌦ ejie⇤i ⌦ e⇤j =

X

i ,j2Ihf , eiejie⇤i ⌦ e⇤j ,

e.g. �E⇤i = ,

10 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

2.) construct the dual Uq(b+)⇤

Calculate the product of f , g 2 Uq(b+)⇤w.r.t. ei 2 {En1

1

En1212

En22

Hm11

Hm22

}

fg =X

i2Ihfg , ei ie⇤i =

X

i2Ihf , (ei )

(1)

ihg , (ei )(2)

ie⇤i ,

e.g.

H⇤i E⇤j = � ~�ij E⇤j

E⇤j H⇤i =

�Ej = Ej ⌦ 1 + q�Hj ⌦ Ej

Calculate the coproduct of f 2 H⇤

�(f ) =X

i ,j2Ih�(f ), ei ⌦ ejie⇤i ⌦ e⇤j =

X

i ,j2Ihf , eiejie⇤i ⌦ e⇤j ,

e.g. �E⇤i = ,

10 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

2.) construct the dual Uq(b+)⇤

Calculate the product of f , g 2 Uq(b+)⇤w.r.t. ei 2 {En1

1

En1212

En22

Hm11

Hm22

}

fg =X

i2Ihfg , ei ie⇤i =

X

i2Ihf , (ei )

(1)

ihg , (ei )(2)

ie⇤i ,

e.g.

H⇤i E⇤j = � ~�ij E⇤j + (EjHi )⇤

E⇤j H⇤i = + (EjHi )

⇤ �EjHi = · · ·+ E⇤j ⌦ Hi + q�Hj Hi ⌦ Ej

Calculate the coproduct of f 2 H⇤

�(f ) =X

i ,j2Ih�(f ), ei ⌦ ejie⇤i ⌦ e⇤j =

X

i ,j2Ihf , eiejie⇤i ⌦ e⇤j ,

e.g. �E⇤i = ,

10 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

2.) construct the dual Uq(b+)⇤

Calculate the product of f , g 2 Uq(b+)⇤w.r.t. ei 2 {En1

1

En1212

En22

Hm11

Hm22

}

fg =X

i2Ihfg , ei ie⇤i =

X

i2Ihf , (ei )

(1)

ihg , (ei )(2)

ie⇤i ,

e.g.

H⇤i E⇤j = � ~�ij E⇤j + (EjHi )⇤

E⇤j H⇤i = + (EjHi )

⇤

Calculate the coproduct of f 2 H⇤

�(f ) =X

i ,j2Ih�(f ), ei ⌦ ejie⇤i ⌦ e⇤j =

X

i ,j2Ihf , eiejie⇤i ⌦ e⇤j ,

e.g. �E⇤i = ,

10 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

2.) construct the dual Uq(b+)⇤

Calculate the product of f , g 2 Uq(b+)⇤w.r.t. ei 2 {En1

1

En1212

En22

Hm11

Hm22

}

fg =X

i2Ihfg , ei ie⇤i =

X

i2Ihf , (ei )

(1)

ihg , (ei )(2)

ie⇤i ,

e.g.

H⇤i E⇤j = � ~�ij E⇤j + (EjHi )⇤

E⇤j H⇤i = + (EjHi )

⇤

Calculate the coproduct of f 2 H⇤

�(f ) =X

i ,j2Ih�(f ), ei ⌦ ejie⇤i ⌦ e⇤j =

X

i ,j2Ihf , eiejie⇤i ⌦ e⇤j ,

e.g. �E⇤i = E⇤i ⌦ 1, Ej1 = Ej

10 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

2.) construct the dual Uq(b+)⇤

Calculate the product of f , g 2 Uq(b+)⇤w.r.t. ei 2 {En1

1

En1212

En22

Hm11

Hm22

}

fg =X

i2Ihfg , ei ie⇤i =

X

i2Ihf , (ei )

(1)

ihg , (ei )(2)

ie⇤i ,

e.g.

H⇤i E⇤j = � ~�ij E⇤j + (EjHi )⇤

E⇤j H⇤i = + (EjHi )

⇤

Calculate the coproduct of f 2 H⇤

�(f ) =X

i ,j2Ih�(f ), ei ⌦ ejie⇤i ⌦ e⇤j =

X

i ,j2Ihf , eiejie⇤i ⌦ e⇤j ,

e.g. �E⇤i = E⇤i ⌦ 1 + e

aij H⇤i ⌦ E⇤i , H

ni Ej = Ej (Hi + aij )

n= anijEj + · · ·

10 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How it works for Uq(sl(3)) [Drinfeld ’86],[Burroughs ’90]1.) positve Borel sub-algebra Uq(b+), basis {E n1

1

E n1212

E n22

Hm11

Hm22

}[Hi ,Ej ] = aijEj , �Ej = Ej ⌦ 1 + q�Hi ⌦ Ej

2.) construct the dual Uq(b+)⇤, dual basis {(E n11

E n1212

E n22

Hm11

Hm22

)⇤}[H⇤i ,E

⇤j ] = �~�ij E⇤j , �E⇤j = E⇤j ⌦ 1 + e

Pi aij H

⇤i ⌦ E⇤j

3.) built the Quantum Double, i.e. find the cross-relations

[Ei ,E⇤j ] = �ij (e

Pk akjH

⇤k � q�Hi )

4.) Identify Uq(b+)⇤op ⇠= Uq(b�)

( issue for psl(2|2)n C3

bHj ⇠1

~X

i

aijH⇤i , Fj ⇠ E

⇤i ,

and thus Uq(sl(3)) ⇠= D(Uq(b+))/h bH � Hi5.) R-matrix eXq :=

P1n=0

Xn

[n;q]! , [n; q] :=1�qn1�q

R = e(q�q�1

)E1

⌦F1

q�2 e(q�q�1)E

12

⌦F21

q�2 e(q�q�1)E

2

⌦F2

q�2 qH1

⌦(2H1

+H2

)/3 qH2⌦(2H2+H1)/3,

11 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How it works for Uq(sl(3)) [Drinfeld ’86],[Burroughs ’90]1.) positve Borel sub-algebra Uq(b+), basis {E n1

1

E n1212

E n22

Hm11

Hm22

}[Hi ,Ej ] = aijEj , �Ej = Ej ⌦ 1 + q�Hi ⌦ Ej

2.) construct the dual Uq(b+)⇤, dual basis {(E n11

E n1212

E n22

Hm11

Hm22

)⇤}[H⇤i ,E

⇤j ] = �~�ij E⇤j , �E⇤j = E⇤j ⌦ 1 + e

Pi aij H

⇤i ⌦ E⇤j

3.) built the Quantum Double, i.e. find the cross-relations

[Ei ,E⇤j ] = �ij (e

Pk akjH

⇤k � q�Hi )

4.) Identify Uq(b+)⇤op ⇠= Uq(b�)

( issue for psl(2|2)n C3

bHj ⇠1

~X

i

aijH⇤i , Fj ⇠ E

⇤i ,

and thus Uq(sl(3)) ⇠= D(Uq(b+))/h bH � Hi5.) R-matrix eXq :=

P1n=0

Xn

[n;q]! , [n; q] :=1�qn1�q

R = e(q�q�1

)E1

⌦F1

q�2 e(q�q�1)E

12

⌦F21

q�2 e(q�q�1)E

2

⌦F2

q�2 qH1

⌦(2H1

+H2

)/3 qH2⌦(2H2+H1)/3,

11 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How it works for Uq(sl(3)) [Drinfeld ’86],[Burroughs ’90]1.) positve Borel sub-algebra Uq(b+), basis {E n1

1

E n1212

E n22

Hm11

Hm22

}[Hi ,Ej ] = aijEj , �Ej = Ej ⌦ 1 + q�Hi ⌦ Ej

2.) construct the dual Uq(b+)⇤, dual basis {(E n11

E n1212

E n22

Hm11

Hm22

)⇤}[H⇤i ,E

⇤j ] = �~�ij E⇤j , �E⇤j = E⇤j ⌦ 1 + e

Pi aij H

⇤i ⌦ E⇤j

3.) built the Quantum Double, i.e. find the cross-relations

[Ei ,E⇤j ] = �ij (e

Pk akjH

⇤k � q�Hi )

4.) Identify Uq(b+)⇤op ⇠= Uq(b�)

( issue for psl(2|2)n C3

bHj ⇠1

~X

i

aijH⇤i , Fj ⇠ E

⇤i ,

and thus Uq(sl(3)) ⇠= D(Uq(b+))/h bH � Hi

5.) R-matrix eXq :=P1

n=0Xn

[n;q]! , [n; q] :=1�qn1�q

R = e(q�q�1

)E1

⌦F1

q�2 e(q�q�1)E

12

⌦F21

q�2 e(q�q�1)E

2

⌦F2

q�2 qH1

⌦(2H1

+H2

)/3 qH2⌦(2H2+H1)/3,

11 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How it works for Uq(sl(3)) [Drinfeld ’86],[Burroughs ’90]1.) positve Borel sub-algebra Uq(b+), basis {E n1

1

E n1212

E n22

Hm11

Hm22

}[Hi ,Ej ] = aijEj , �Ej = Ej ⌦ 1 + q�Hi ⌦ Ej

2.) construct the dual Uq(b+)⇤, dual basis {(E n11

E n1212

E n22

Hm11

Hm22

)⇤}[H⇤i ,E

⇤j ] = �~�ij E⇤j , �E⇤j = E⇤j ⌦ 1 + e

Pi aij H

⇤i ⌦ E⇤j

3.) built the Quantum Double, i.e. find the cross-relations

[Ei ,E⇤j ] = �ij (e

Pk akjH

⇤k � q�Hi )

4.) Identify Uq(b+)⇤op ⇠= Uq(b�)

( issue for psl(2|2)n C3

bHj ⇠1

~X

i

aijH⇤i , Fj ⇠ E

⇤i ,

and thus Uq(sl(3)) ⇠= D(Uq(b+))/h bH � Hi5.) R-matrix eXq :=

P1n=0

Xn

[n;q]! , [n; q] :=1�qn1�q

R = e(q�q�1

)E1

⌦F1

q�2 e(q�q�1)E

12

⌦F21

q�2 e(q�q�1)E

2

⌦F2

q�2 qH1

⌦(2H1

+H2

)/3 qH2⌦(2H2+H1)/3,

11 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How it works for Uq(sl(3)) [Drinfeld ’86],[Burroughs ’90]1.) positve Borel sub-algebra Uq(b+), basis {E n1

1

E n1212

E n22

Hm11

Hm22

}[Hi ,Ej ] = aijEj , �Ej = Ej ⌦ 1 + q�Hi ⌦ Ej

2.) construct the dual Uq(b+)⇤, dual basis {(E n11

E n1212

E n22

Hm11

Hm22

)⇤}[H⇤i ,E

⇤j ] = �~�ij E⇤j , �E⇤j = E⇤j ⌦ 1 + e

Pi aij H

⇤i ⌦ E⇤j

3.) built the Quantum Double, i.e. find the cross-relations

[Ei ,E⇤j ] = �ij (e

Pk akjH

⇤k � q�Hi )

4.) Identify Uq(b+)⇤op ⇠= Uq(b�) ( issue for psl(2|2)n C3

bHj ⇠1

~X

i

aijH⇤i , Fj ⇠ E

⇤i ,

and thus Uq(sl(3)) ⇠= D(Uq(b+))/h bH � Hi5.) R-matrix eXq :=

P1n=0

Xn

[n;q]! , [n; q] :=1�qn1�q

R = e(q�q�1

)E1

⌦F1

q�2 e(q�q�1)E

12

⌦F21

q�2 e(q�q�1)E

2

⌦F2

q�2 qH1

⌦(2H1

+H2

)/3 qH2⌦(2H2+H1)/3,

11 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Quantum Double for sl(2|2)

12 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Now for Uq(psl(2|2)n C3)cartan sub-algebra

[H⇤i ,E⇤j ] =

a⇤ijz }| {�~�ijE ⇤j

dual Cartan matrix a⇤ij is not degenerate

Serre relations

E ⇤1

. E ⇤3

= 0, E ⇤i . (E⇤i . E

⇤2

) = 0, {E ⇤12

,E ⇤32

} = 0

no central extension from quartic Serre relation

Instead non central P⇤

[H⇤i ,P⇤] = �~(�i1 + 2�2i + �i3)P⇤,

⇥P⇤,E⇤j

⇤= �j3

�q � q�1

�E⇤32

E⇤132

.

P⇤ not central

) no chance to identify: Uq(b+)⇤op � Uq(b�)

13 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Now for Uq(psl(2|2)n C3)cartan sub-algebra

[H⇤i ,E⇤j ] =

a⇤ijz }| {�~�ijE ⇤j

dual Cartan matrix a⇤ij is not degenerate

Serre relations

E ⇤1

. E ⇤3

= 0, E ⇤i . (E⇤i . E

⇤2

) = 0, {E ⇤12

,E ⇤32

} = 0

no central extension from quartic Serre relation

Instead non central P⇤

[H⇤i ,P⇤] = �~(�i1 + 2�2i + �i3)P⇤,

⇥P⇤,E⇤j

⇤= �j3

�q � q�1

�E⇤32

E⇤132

.

P⇤ not central

) no chance to identify: Uq(b+)⇤op � Uq(b�)

13 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Now for Uq(psl(2|2)n C3)cartan sub-algebra

[H⇤i ,E⇤j ] =

a⇤ijz }| {�~�ijE ⇤j

dual Cartan matrix a⇤ij is not degenerate

Serre relations

E ⇤1

. E ⇤3

= 0, E ⇤i . (E⇤i . E

⇤2

) = 0, {E ⇤12

,E ⇤32

} = 0

no central extension from quartic Serre relation

Instead non central P⇤

[H⇤i ,P⇤] = �~(�i1 + 2�2i + �i3)P⇤,

⇥P⇤,E⇤j

⇤= �j3

�q � q�1

�E⇤32

E⇤132

.

P⇤ not central

) no chance to identify: Uq(b+)⇤op � Uq(b�)

13 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Now for Uq(psl(2|2)n C3)cartan sub-algebra

[H⇤i ,E⇤j ] =

a⇤ijz }| {�~�ijE ⇤j

dual Cartan matrix a⇤ij is not degenerate

Serre relations

E ⇤1

. E ⇤3

= 0, E ⇤i . (E⇤i . E

⇤2

) = 0, {E ⇤12

,E ⇤32

} = 0

no central extension from quartic Serre relation

Instead non central P⇤

[H⇤i ,P⇤] = �~(�i1 + 2�2i + �i3)P⇤,

⇥P⇤,E⇤j

⇤= �j3

�q � q�1

�E⇤32

E⇤132

.

P⇤ not central

) no chance to identify: Uq(b+)⇤op � Uq(b�)13 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Overview

Uq(b+)

Uq(b+)⇤

Uq(b�)

psl(2|2) H1/3

H⇤1/3

H1/3

Ei

E ⇤i

Fi

C3 C

H⇤A

C

P

E ⇤A

K

sl(2) HA

C ⇤ HA

EA

P⇤ FA

Solution: Enlarge Hopf algebra with additional generatorsEA

,HA

,FA

14 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Overview

Uq(b+) Uq(b+)⇤ Uq(b�)

psl(2|2) H1/3

⇤ // H⇤1/3

⇠ H1/3

Ei⇤ // E ⇤i

⇠ Fi

C3 C⇤

%%

H⇤A

C

P⇤

%%

E ⇤A

K

sl(2) HA

C ⇤

HA

EA

P⇤

FA

Solution: Enlarge Hopf algebra with additional generatorsEA

,HA

,FA

14 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Overview

Uq(b+) Uq(b+)⇤ Uq(b�)

psl(2|2) H1/3

⇤ // H⇤1/3

⇠ H1/3

Ei⇤ // E ⇤i

⇠ Fi

C3 C⇤

%%

H⇤A

⇠ C

P⇤

%%

E ⇤A

⇠ K

sl(2) HA

⇤99

C ⇤ ⇠ HA

EA

⇤99

P⇤ ⇠ FA

Solution: Enlarge Hopf algebra with additional generatorsEA

,HA

,FA

14 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How to find Hopf algebra of EA, HA, K?

add HA

, s.t its dual H⇤A

is central (in psl(2|2)n C3).

[H⇤A

,E ⇤j ] = 0,[H⇤

A

,F ⇤j ] = 0,�H⇤

A

= H⇤A

⌦ 1 + 1⌦ H⇤A

.

add EA

, s.t. E ⇤A

is central (in psl(2|2)n C3) and

{E ⇤12

,E ⇤32

} = E ⇤A

, �E ⇤A

= E ⇤A

⌦ 1 + e2H⇤A ⌦ E ⇤A

add FA

, as given by P⇤

requireUq(b+)⇤op ⇠= Uq(b�)

) one-parameter (⇠) family of consistent Hopf algebras

15 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How to find Hopf algebra of EA, HA, K?

add HA

, s.t its dual H⇤A

is central (in psl(2|2)n C3).

[H⇤A

,E ⇤j ] = 0,[H⇤

A

,F ⇤j ] = 0,�H⇤

A

= H⇤A

⌦ 1 + 1⌦ H⇤A

.

add EA

, s.t. E ⇤A

is central (in psl(2|2)n C3) and

{E ⇤12

,E ⇤32

} = E ⇤A

, �E ⇤A

= E ⇤A

⌦ 1 + e2H⇤A ⌦ E ⇤A

add FA

, as given by P⇤

requireUq(b+)⇤op ⇠= Uq(b�)

) one-parameter (⇠) family of consistent Hopf algebras

15 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How to find Hopf algebra of EA, HA, K?

add HA

, s.t its dual H⇤A

is central (in psl(2|2)n C3).

[H⇤A

,E ⇤j ] = 0,[H⇤

A

,F ⇤j ] = 0,�H⇤

A

= H⇤A

⌦ 1 + 1⌦ H⇤A

.

add EA

, s.t. E ⇤A

is central (in psl(2|2)n C3) and

{E ⇤12

,E ⇤32

} = E ⇤A

, �E ⇤A

= E ⇤A

⌦ 1 + e2H⇤A ⌦ E ⇤A

add FA

, as given by P⇤

requireUq(b+)⇤op ⇠= Uq(b�)

) one-parameter (⇠) family of consistent Hopf algebras

15 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How to find Hopf algebra of EA, HA, K?

add HA

, s.t its dual H⇤A

is central (in psl(2|2)n C3).

[H⇤A

,E ⇤j ] = 0,[H⇤

A

,F ⇤j ] = 0,�H⇤

A

= H⇤A

⌦ 1 + 1⌦ H⇤A

.

add EA

, s.t. E ⇤A

is central (in psl(2|2)n C3) and

{E ⇤12

,E ⇤32

} = E ⇤A

, �E ⇤A

= E ⇤A

⌦ 1 + e2H⇤A ⌦ E ⇤A

add FA

, as given by P⇤

requireUq(b+)⇤op ⇠= Uq(b�)

) one-parameter (⇠) family of consistent Hopf algebras

15 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

How to find Hopf algebra of EA, HA, K?

add HA

, s.t its dual H⇤A

is central (in psl(2|2)n C3).

[H⇤A

,E ⇤j ] = 0,[H⇤

A

,F ⇤j ] = 0,�H⇤

A

= H⇤A

⌦ 1 + 1⌦ H⇤A

.

add EA

, s.t. E ⇤A

is central (in psl(2|2)n C3) and

{E ⇤12

,E ⇤32

} = E ⇤A

, �E ⇤A

= E ⇤A

⌦ 1 + e2H⇤A ⌦ E ⇤A

add FA

, as given by P⇤

requireUq(b+)⇤op ⇠= Uq(b�)

) one-parameter (⇠) family of consistent Hopf algebras

15 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Hopf structure of sl(2) automorphism

sl(2)

[HA

,EA

] = 2EA

, [HA

,FA

] = �2FA

, [EA

,FA

] = (HA

+ ⇠C)q2C + q�2C

2

� ⇠q2C � q�2C

4~,

acting on central extensions

[HA

,P] = 2P, [HA

,C ] = 0, [HA

,K ] = �2K ,

[EA

,P] = 0, [EA

,C ] =q � q�1

2~P, [E

A

,K ] =q2C � q�2C

q � q�1,

[FA

,P] = �q2C � q�2C

q � q�1, [F

A

,C ] = �q � q�1

2~K , [F

A

,K ] = 0,

acting on psl(2|2)[H

A

,Ej ] = �j2E2, [HA,Fj ] = ��j2Fj[E

A

,Ej ] = (q � q�1)(�j2 12

E2

P + �j3E32E132),

[EA

,Fj ] = �j2(E132 + (q � q�1)E32E1)q�H2 + �j3(q � q�1)E2E12qH3 ,E $ F , P $ K ,

16 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Hopf structure of sl(2) automorphism

sl(2)

[HA

,EA

] = 2EA

, [HA

,FA

] = �2FA

, [EA

,FA

] = (HA

+ ⇠C)q2C + q�2C

2

� ⇠q2C � q�2C

4~,

acting on central extensions

[HA

,P] = 2P, [HA

,C ] = 0, [HA

,K ] = �2K ,

[EA

,P] = 0, [EA

,C ] =q � q�1

2~P, [E

A

,K ] =q2C � q�2C

q � q�1,

[FA

,P] = �q2C � q�2C

q � q�1, [F

A

,C ] = �q � q�1

2~K , [F

A

,K ] = 0,

acting on psl(2|2)[H

A

,Ej ] = �j2E2, [HA,Fj ] = ��j2Fj[E

A

,Ej ] = (q � q�1)(�j2 12

E2

P + �j3E32E132),

[EA

,Fj ] = �j2(E132 + (q � q�1)E32E1)q�H2 + �j3(q � q�1)E2E12qH3 ,E $ F , P $ K ,

16 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Hopf structure of sl(2) automorphism

sl(2)

[HA

,EA

] = 2EA

, [HA

,FA

] = �2FA

, [EA

,FA

] = (HA

+ ⇠C)q2C + q�2C

2

� ⇠q2C � q�2C

4~,

acting on central extensions

[HA

,P] = 2P, [HA

,C ] = 0, [HA

,K ] = �2K ,

[EA

,P] = 0, [EA

,C ] =q � q�1

2~P, [E

A

,K ] =q2C � q�2C

q � q�1,

[FA

,P] = �q2C � q�2C

q � q�1, [F

A

,C ] = �q � q�1

2~K , [F

A

,K ] = 0,

acting on psl(2|2)[H

A

,Ej ] = �j2E2, [HA,Fj ] = ��j2Fj[E

A

,Ej ] = (q � q�1)(�j2 12

E2

P + �j3E32E132),

[EA

,Fj ] = �j2(E132 + (q � q�1)E32E1)q�H2 + �j3(q � q�1)E2E12qH3 ,E $ F , P $ K ,

16 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Hopf structure of sl(2) automorphisms

coproduct

�HA

= HA

⌦ 1 + 1⌦ HA

�EA

= EA

⌦ 1 + q�2C ⌦ EA

+q � q�1

2(H

A

+ ⇠C )q�2C ⌦ P

+ (q � q�1)(E132

+ (q � q�1)E32

E1

)q�H2 ⌦ E2

� (q � q�1)E32

q�H1�H2 ⌦ E12

� (q � q�1)2E3

q�H1�2H2 ⌦ E2

E12

�FA

= analogously

17 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R =1X

n1

,··· ,nl

�en11

· · · enll�⌦�en11

· · · enll�⇤

=1X

n1

en11

⌦ e⇤n11

he⇤n11

, en11

i · · ·1X

nl

enll ⌦ e⇤nll

he⇤nll , en1

l i.

provided e⇤n11

· · · e⇤nll = he⇤n

1

1

, en11

i · · · he⇤nll , enll i

�en11

· · · enll�⇤

.

) choose your basis wisely!! Problem: sl(2) and C3 generators do not factorize

R = e�E2⌦F2e�E12⌦F21

· exp✓

EA

⌦ K + P ⌦ FA

�P ⌦ K �⇠

2~

◆log(1 + �2P ⌦ K )� ⇠

4~ Li2(��2P ⌦ K )

�

·e�E32⌦F32e�E132⌦F132e�E1⌦F1q�2 e��E

3

⌦F3

q2 q1

2

H1

⌦H1

� 12

H3

⌦H3

+C⌦HA

+HA

⌦C+ ⇠C⌦C

� = (q � q�1)

18 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R =1X

n1

,··· ,nl

�en11

· · · enll�⌦�en11

· · · enll�⇤

=1X

n1

en11

⌦ e⇤n11

he⇤n11

, en11

i · · ·1X

nl

enll ⌦ e⇤nll

he⇤nll , en1

l i.

provided e⇤n11

· · · e⇤nll = he⇤n

1

1

, en11

i · · · he⇤nll , enll i

�en11

· · · enll�⇤

.

) choose your basis wisely!! Problem: sl(2) and C3 generators do not factorize

R = e�E2⌦F2e�E12⌦F21

· exp✓

EA

⌦ K + P ⌦ FA

�P ⌦ K �⇠

2~

◆log(1 + �2P ⌦ K )� ⇠

4~ Li2(��2P ⌦ K )

�

·e�E32⌦F32e�E132⌦F132e�E1⌦F1q�2 e��E

3

⌦F3

q2 q1

2

H1

⌦H1

� 12

H3

⌦H3

+C⌦HA

+HA

⌦C+ ⇠C⌦C

� = (q � q�1)

18 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R =1X

n1

,··· ,nl

�en11

· · · enll�⌦�en11

· · · enll�⇤

=1X

n1

en11

⌦ e⇤n11

he⇤n11

, en11

i · · ·1X

nl

enll ⌦ e⇤nll

he⇤nll , en1

l i.

provided e⇤n11

· · · e⇤nll = he⇤n

1

1

, en11

i · · · he⇤nll , enll i

�en11

· · · enll�⇤

.

) choose your basis wisely!

! Problem: sl(2) and C3 generators do not factorize

R = e�E2⌦F2e�E12⌦F21

· exp✓

EA

⌦ K + P ⌦ FA

�P ⌦ K �⇠

2~

◆log(1 + �2P ⌦ K )� ⇠

4~ Li2(��2P ⌦ K )

�

·e�E32⌦F32e�E132⌦F132e�E1⌦F1q�2 e��E

3

⌦F3

q2 q1

2

H1

⌦H1

� 12

H3

⌦H3

+C⌦HA

+HA

⌦C+ ⇠C⌦C

� = (q � q�1)

18 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R =1X

n1

,··· ,nl

�en11

· · · enll�⌦�en11

· · · enll�⇤

=1X

n1

en11

⌦ e⇤n11

he⇤n11

, en11

i · · ·1X

nl

enll ⌦ e⇤nll

he⇤nll , en1

l i.

provided e⇤n11

· · · e⇤nll = he⇤n

1

1

, en11

i · · · he⇤nll , enll i

�en11

· · · enll�⇤

.

) choose your basis wisely!! Problem: sl(2) and C3 generators do not factorize

R = e�E2⌦F2e�E12⌦F21

· exp✓

EA

⌦ K + P ⌦ FA

�P ⌦ K �⇠

2~

◆log(1 + �2P ⌦ K )� ⇠

4~ Li2(��2P ⌦ K )

�

·e�E32⌦F32e�E132⌦F132e�E1⌦F1q�2 e��E

3

⌦F3

q2 q1

2

H1

⌦H1

� 12

H3

⌦H3

+C⌦HA

+HA

⌦C+ ⇠C⌦C

� = (q � q�1)

18 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R =1X

n1

,··· ,nl

�en11

· · · enll�⌦�en11

· · · enll�⇤

=1X

n1

en11

⌦ e⇤n11

he⇤n11

, en11

i · · ·1X

nl

enll ⌦ e⇤nll

he⇤nll , en1

l i.

provided e⇤n11

· · · e⇤nll = he⇤n

1

1

, en11

i · · · he⇤nll , enll i

�en11

· · · enll�⇤

.

) choose your basis wisely!! Problem: sl(2) and C3 generators do not factorize

R = e�E2⌦F2e�E12⌦F21

· exp✓

EA

⌦ K + P ⌦ FA

�P ⌦ K �⇠

2~

◆log(1 + �2P ⌦ K )� ⇠

4~ Li2(��2P ⌦ K )

�

·e�E32⌦F32e�E132⌦F132e�E1⌦F1q�2 e��E

3

⌦F3

q2 q1

2

H1

⌦H1

� 12

H3

⌦H3

+C⌦HA

+HA

⌦C+ ⇠C⌦C

� = (q � q�1)18 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum double

extended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum DoubleUnusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum doubleextended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum DoubleUnusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum doubleextended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum Double

Unusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum doubleextended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum DoubleUnusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum doubleextended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum DoubleUnusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum doubleextended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum DoubleUnusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum doubleextended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum DoubleUnusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum doubleextended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum DoubleUnusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

So far

cannot write Uq(psl(2|2)n C3) as a quantum doubleextended with sl(2) automorphisms and found Hopf algebra

Uq,⇠(sl(2)n psl(2|2)n C3) is a Quantum DoubleUnusual features

[EA

,FA

] = (HA

+ ⇠C )q2C + q�2C

2� ⇠ q

2C � q�2C4~

sl(2) automorphism not deformed in the usual way

appearance of plain ~ instead of q = e~

appearance of free parameter ⇠

R-matrix does not factorize and appearance of log and Li2

Can we understand their origin better?

19 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Contraction Limit

20 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Plan:Focus on Uq,⇠(sl(2)n C3)re-derive it as a limit ✏, ✏̃ ! 0 of Uq✏(sl(2))⌦ Uq✏̃(sl(2))

Uq✏(sl(2))[H,E ] = 2E ,[H,F ] = �2F , [E ,F ] =

q✏H � q�✏Hq✏ � q�✏

�E = E ⌦ 1 + q�✏H ⌦ E ,�F = F ⌦ q✏H + 1⌦ F , �H = H ⌦ 1 + 1⌦ H.

Define

EA

:= E + Ẽ , P := ✏E ,

FA

:= F + F̃ , K := ✏F ,

HA

:= H + H̃, 2C := ✏H.

stay sl(2), ”loose” q, commute , ”keep” q,

21 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Plan:Focus on Uq,⇠(sl(2)n C3)re-derive it as a limit ✏, ✏̃ ! 0 of Uq✏(sl(2))⌦ Uq✏̃(sl(2))

Uq✏(sl(2))[H,E ] = 2E ,[H,F ] = �2F , [E ,F ] =

q✏H � q�✏Hq✏ � q�✏

�E = E ⌦ 1 + q�✏H ⌦ E ,�F = F ⌦ q✏H + 1⌦ F , �H = H ⌦ 1 + 1⌦ H.

Define

EA

:= E + Ẽ , P := ✏E ,

FA

:= F + F̃ , K := ✏F ,

HA

:= H + H̃, 2C := ✏H.

stay sl(2), ”loose” q, commute , ”keep” q,

21 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Plan:Focus on Uq,⇠(sl(2)n C3)re-derive it as a limit ✏, ✏̃ ! 0 of Uq✏(sl(2))⌦ Uq✏̃(sl(2))

Uq✏(sl(2))[H,E ] = 2E ,[H,F ] = �2F , [E ,F ] =

q✏H � q�✏Hq✏ � q�✏

�E = E ⌦ 1 + q�✏H ⌦ E ,�F = F ⌦ q✏H + 1⌦ F , �H = H ⌦ 1 + 1⌦ H.

Define

EA

:= E + Ẽ , P := ✏E ,

FA

:= F + F̃ , K := ✏F ,

HA

:= H + H̃, 2C := ✏H.

stay sl(2), ”loose” q, commute , ”keep” q,

21 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Plan:Focus on Uq,⇠(sl(2)n C3)re-derive it as a limit ✏, ✏̃ ! 0 of Uq✏(sl(2))⌦ Uq✏̃(sl(2))

Uq✏(sl(2))[H,E ] = 2E ,[H,F ] = �2F , [E ,F ] =

q✏H � q�✏Hq✏ � q�✏

�E = E ⌦ 1 + q�✏H ⌦ E ,�F = F ⌦ q✏H + 1⌦ F , �H = H ⌦ 1 + 1⌦ H.

Define

EA

:= E + Ẽ , P := ✏E ,

FA

:= F + F̃ , K := ✏F ,

HA

:= H + H̃, 2C := ✏H.

stay sl(2), ”loose” q, commute , ”keep” q,21 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Limit

Take simultaneous limit ✏, ✏̃ ! 0

✏̃(✏) = �✏+ 12

⇠✏2 +O(✏3).

Check for divergences

�EA

=1

✏

�q�2C � q�2C

�⌦ P +O(✏0),

) � = �1No constraint on ⇠

Recover Uq,⇠(sl(2)n C3)

22 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Limit

Take simultaneous limit ✏, ✏̃ ! 0

✏̃(✏) = �✏+ 12

⇠✏2 +O(✏3).

Check for divergences

�EA

=1

✏

�q�2C � q�2C

�⌦ P +O(✏0),

) � = �1

No constraint on ⇠

Recover Uq,⇠(sl(2)n C3)

22 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Limit

Take simultaneous limit ✏, ✏̃ ! 0

✏̃(✏) = �✏+ 12

⇠✏2 +O(✏3).

Check for divergences

�EA

=1

✏

�q�2C � q�2C

�⌦ P +O(✏0),

) � = �1No constraint on ⇠

Recover Uq,⇠(sl(2)n C3)

22 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Limit

Take simultaneous limit ✏, ✏̃ ! 0

✏̃(✏) = �✏+ 12

⇠✏2 +O(✏3).

Check for divergences

�EA

=1

✏

�q�2C � q�2C

�⌦ P +O(✏0),

) � = �1No constraint on ⇠

Recover Uq,⇠(sl(2)n C3)

22 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R = e(q✏�q�✏)E⌦F

q�2✏ q1

2

✏H⌦H e(q✏̃�q�✏̃) ˜E⌦ ˜F

q�2✏̃q1

2

✏̃ ˜H⌦ ˜H

= expq�2✏

�1✏P ⌦ K

�expq�2✏̃

✏̃

✏2(P + ✏E

A

)⌦ (K + ✏FA

)

�

· q 2✏C⌦Cq✏̃

2✏2(2C�✏H

A

)⌦(2C�✏HA

)

q-Dilogarithm: Li2

[X ; q] := log expq[X ]

Li2

X (✏)

✏; q✏

�= � 1~✏ Li2

��~X (✏)

�+O(✏).

) R-matrix of Uq,⇠(sl(2)n C3) reproduced

23 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R = e(q✏�q�✏)E⌦F

q�2✏ q1

2

✏H⌦H e(q✏̃�q�✏̃) ˜E⌦ ˜F

q�2✏̃q1

2

✏̃ ˜H⌦ ˜H

= expq�2✏

�1✏P ⌦ K

�expq�2✏̃

✏̃

✏2(P + ✏E

A

)⌦ (K + ✏FA

)

�

· q 2✏C⌦Cq✏̃

2✏2(2C�✏H

A

)⌦(2C�✏HA

)

q-Dilogarithm: Li2

[X ; q] := log expq[X ]

Li2

X (✏)

✏; q✏

�= � 1~✏ Li2

��~X (✏)

�+O(✏).

) R-matrix of Uq,⇠(sl(2)n C3) reproduced

23 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R = e(q✏�q�✏)E⌦F

q�2✏ q1

2

✏H⌦H e(q✏̃�q�✏̃) ˜E⌦ ˜F

q�2✏̃q1

2

✏̃ ˜H⌦ ˜H

= expq�2✏

�1✏P ⌦ K

�expq�2✏̃

✏̃

✏2(P + ✏E

A

)⌦ (K + ✏FA

)

�

· q 2✏C⌦Cq✏̃

2✏2(2C�✏H

A

)⌦(2C�✏HA

)

q-Dilogarithm: Li2

[X ; q] := log expq[X ]

Li2

X (✏)

✏; q✏

�= � 1~✏ Li2

��~X (✏)

�+O(✏).

) R-matrix of Uq,⇠(sl(2)n C3) reproduced

23 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

R-matrix

R = e(q✏�q�✏)E⌦F

q�2✏ q1

2

✏H⌦H e(q✏̃�q�✏̃) ˜E⌦ ˜F

q�2✏̃q1

2

✏̃ ˜H⌦ ˜H

= expq�2✏

�1✏P ⌦ K

�expq�2✏̃

✏̃

✏2(P + ✏E

A

)⌦ (K + ✏FA

)

�

· q 2✏C⌦Cq✏̃

2✏2(2C�✏H

A

)⌦(2C�✏HA

)

q-Dilogarithm: Li2

[X ; q] := log expq[X ]

Li2

X (✏)

✏; q✏

�= � 1~✏ Li2

��~X (✏)

�+O(✏).

) R-matrix of Uq,⇠(sl(2)n C3) reproduced

23 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Last words

”undeformed” sl(2) and plain ~ from q✏ = e✏~

free parameter ⇠ stems from ✏̃(✏) = �✏+ 12

⇠✏2

dilogarithm in R-matrix from log expq

Uq,⇠(sl(2)n C3) with ⇠ = 0 known as 3D -PoincaréTo include psl(2|2) start with Uq(d(2, 1; ✏))⌦ Uq✏̃(sl(2))Also possible Uq(d(2, 1; ✏))⌦ Uq(d(2, 1; ✏̃))! Uq,⇠(sl(2)n (psl(2|2)� psl(2|2))n C3)

24 / 24

psl(2|2) n C3 Quantum Double Quantum Double for sl(2|2) Contraction Limit

Last words

”undeformed” sl(2) and plain ~ from q✏ = e✏~

free parameter ⇠ stems from ✏̃(✏) = �✏+ 12

⇠✏2

dilogarithm in R-matrix from log expq

Uq,⇠(sl(2)n C3) with ⇠ = 0 known as 3D -PoincaréTo include psl(2|2) start with Uq(d(2, 1; ✏))⌦ Uq✏̃(sl(2))Also possible Uq(d(2, 1; ✏))⌦ Uq(d(2, 1; ✏̃))! Uq,⇠(sl(2)n (psl(2|2)� psl(2|2))n C3)

24 / 24

Centrally extended sl(2|2)Quantum DoubleQuantum Double for sl(2|2)Contraction Limit