Appendix C. Pentagon and superpentagon relation 125
Appendix C. Pentagon and superpentagon relation 126 Supersymmetric case : In order to prove the Ramanujan formulas in the supersym-metric case , let us define
Φ+NS(a, w) = eNS(a)eNS(w)
eNS(a+w−cb), Φ−NS(a, w) = eNS(a)eR(w) eR(a+w−cb), Φ+R(a, w) = eR(a)eNS(w)
eR(a+w−cb), Φ−R(a, w) = eR(a)eR(w) eNS(a+w−cb).
Ψ+NS(a, w) = Z
dxeπix(w−cb)
eNS(x+a)
eNS(x−cb) + eR(x+a) eb(x−cR)
, Ψ−NS(a, w) =
Z
dxeπix(w−cb)
eNS(x+a)
eNS(x−cb) − eR(x+a) eb(x−cR)
, Ψ+R(a, w) =
Z
dxeπix(w−cb)
eR(x+a)
eNS(x−cb) +eNS(x+a) eb(x−cR)
, Ψ−R(a, w) =
Z
dxeπix(w−cb)
− eR(x+a)
eNS(x−cb)+ eNS(x+a) eb(x−cR)
,
One can consider the shift equations (omitting intermediate steps) Ψ±NS(a−ib
2, w) =±Ψ±R(a+ib
2, w)±ieπabΨ∓R(a+ib
2, w−ib), Ψ±R(a+ib
2, w) + Ψ∓R(a+ ib
2, w−ib) =eπb(w−cb)Ψ±NS(a−ib 2, w), Ψ±R(a−ib
2, w) =±Ψ+NS(a+ib
2, w)±ieπabΨ∓NS(a+ib
2, w−ib), Ψ±NS(a+ib
2, w)−Ψ∓NS(a+ ib
2, w−ib) =eπb(w−cb)Ψ±R(a−ib 2, w).
since
e−1NS(x−cb) = (1−ieπb(x−cb+ib/2))−1e−1R (x−cb+ib) = (1−eπbx)−1e−1R (x−cb+ib), e−1R (x−cb) = (1 +ieπb(x−cb+ib/2))−1e−1NS(x−cb+ib) = (1 +eπbx)−1e−1NS(x−cb+ib).
So, eventually
(1∓ieπb(w+a−cb))Ψ+NS/R(a−ib
2, w) = (1∓ieπab)Ψ+R/NS(a+ib 2, w), (1±ieπb(w+a−cb))Ψ−NS/R(a−ib
2, w) =−(1−ieπab)Ψ−R/NS(a+ib 2, w), (1∓ieπb(w+a−cb))Ψ∓R(a+ib
2, w−ib) =−(1∓eπ(w−cb))Ψ±R(a+ib 2, w), (1±ieπb(w+a−cb))Ψ∓NS(a+ib
2, w−ib) = (1∓eπ(w−cb))Ψ±NS(a+ib 2, w),
Appendix C. Pentagon and superpentagon relation 127 On the other hand, it is easy to check that
Φ+NS/R(a−ib
2, w) = 1∓ieπab
1∓ieπb(a+w−cb)Φ+R/NS(a+ib 2, w), Φ−NS/R(a−ib
2, w) =− 1∓ieπab
1±ieπb(a+w−cb)Φ−R/NS(a+ib 2, w), Φ±NS(a+ib
2, w−ib) = 1±eπb(w−cb)
1∓ieπb(a+w−cb)Φ∓NS(a+ ib 2, w), Φ±R(a+ib
2, w−ib) =− 1±eπb(w−cb)
1±ieπb(a+w−cb)Φ∓R(a+ ib 2, w),
Taking b → b−1 one gets an additional set of shift equations. Given that one has two doubly periodic functions with the same equations, they have to be equal up to a constant:
Ψ±NS(a, w) =CΦ±NS(a, w), Ψ±R(a, w) =CΦ±R(a, w), Then, one can fixC by evaluating the expression on particulara andw.
Moreover, on can use the connection between eR, eNS and GR, GNS to prove the Ra-manujan formulae based on the results from [80].
LHS = Z
R
eNS(x+u)
eNS(x+v) +eR(x+u) eR(x+v)
eπiwxdx=
= Z
GNS(−i(x+v) +Q/2)
GNS(−i(x+u) +Q/2)+ GR(−i(x+v) +Q/2) GR(−i(x+u) +Q/2)
eπiwxdx=
= Z
iR
GNS(τ −iv+Q/2)
GNS(τ−iu+Q/2)+GR(τ−iv+Q/2) GR(τ −iu+Q/2)
eπiτ(iw)dτ i =
= Z
iR
GNS(τ+ (Q/2−iv) + (Q/2 +iu))
GNS(τ +Q) +GR(τ + (Q/2−iv) + (Q/2 +iu)) GR(τ +Q)
×
×eπiτ(iw)eπi(Q/2+iu)(iw)dτ i =
= 2e−iπc2b/2GNS(iw)GNS(Q+iu−iv)
GNS(Q+iu−iv+iw) e−πw(Q/2+iu)
=
= 2e−iπc2b/2A4 A2
eNS(iQ−u+v−w−cb)
eNS(−w−cb)eNS(iQ−u+v−cb)e−πw(Q/2+iu)
=
= 2e−iπc2b/2A2 eNS(−u+v−w+cb)
eNS(−w−cb)eNS(−u+v+cb)e−πw(Q/2+iu)=
= eNS(−u+v−w+cb)
eNS(−w−cb)eNS(−u+v+cb)e−πiw(u−cb)
2e−iπc2b/2A2
=RHS, where
B¯ = 2e−iπc2b/2A2, B = 2eiπc2b/2A−2, Three additional equations can be proven in the same way.
Appendix D
Permutation
In section 6.2.3 we mentioned that the Pf in equation (6.24) is not the standard per-mutation operator on C1|1 ⊗C1|1 satisfying Pf(η1 ⊗ η2)Pf = η2 ⊗η1 for arbitrary η1, η2 ∈ End(C1|1). In this appendix we want to show how to calculate the standard permutation operator onC⊗Cin the supercase.
One can use the Pauli matrices as the bases : Hom (C1|1) = span(σi) =span(I, σi) σ1= 0 1
1 0
!
, σ2 = 0 −i i 0
!
, σ3 = 1 0 0 −1
!
. (D.1)
We consider permutation as
P(12) = Σaijσi⊗σj (D.2)
while the bases have grading as : |σ1 |=|σ4 |= 0,|σ2|=|σ3 |= 1,(−1)δi,2 = (−1)|i| and satisfy the following relations:
σ1ei =ei, σ2ei =ei+1, σ3ei =i(−1)δi,2ei+1, σ4ei = (−1)δi,2ei, We can choose the basis for calculatingP(12) depending on the Pauli’s matrices as
σ1=I, σ2 =σ1, σ3 =σ2, σ4=σ3. The permutation can be calculated as
Σaijσi⊗σj = Σklakl(σk⊗σl)(ei⊗ej) = (−1)|i||j|(ej⊗ei) i, j= 1,2
129
Appendix D. Permutation 130 then we get:
Σklakl(σk.ei)⊗(σl.ej)(−1)|ei||σl|=a11(ei⊗ej) +a12(ei⊗ej+ 1)(−1)|ei|+
a13(ei⊗i(−1)|j|ej+1)(−1)|ei|+a14(ei⊗(−1)|j|ej) +a21(ei+1⊗ej) +a22(ei+1⊗ej+1)(−1)|ei|+ a23(ei+1⊗i(−1)|j|ej+1)(−1)|ei|+a24(ei+1⊗(−1)|j|ej) +a31((−1)|i|iei+1⊗ej)+
a32((−1)|i|iei+1⊗ej+1)(−1)|ei|+a33((−1)|i|iei+1⊗i(−1)|j|ej+1)(−1)|ei|+
a34((−1)|i|iei+1⊗(−1)|j|ej) +a41((−1)|i|iei⊗ej) +a42((−1)|i|ei⊗ej+1)(−1)|ei|+ a43((−1)|i|ei⊗(−1)|j|iej+1)(−1)|ei|+a44((−1)|i|ei⊗(−1)|j|ej)
we have 16 equations for different choices of i and j. By solving these equations we get the following non zero coefficients: a14= 1/2, a23=i/2, a32=−i/2, a41= 1/2.
Then, the permutation it found to be
2P(12) =σ1⊗σ4+iσ2⊗σ3−iσ3⊗σ2+σ4⊗σ1
=I⊗ 1 0 0 −1
!
+ 1 0
0 −1
!
⊗I+ 0 −1
1 0
!
⊗ 0 1 1 0
!
− 0 1 1 0
!
⊗ 0 −1
1 0
! . By knowing that
a b c d
!
⊗ α β γ δ
!
=
aα aβ bα bβ
aγ aδ bγ bδ
cα −cβ dα −bβ
−cγ cδ −dγ bδ
. (D.3)
Then permutation is
P(12) =
1 0 0 0
0 0 −1 0
0 −1 0 0
0 0 0 −1
, (D.4)
and satisfies the necessary properties P(12)e1⊗e1=e1⊗e1,
P(12)e2⊗e2=e2⊗e2, P(12)2= 1, P(12)e3⊗e3=e3⊗e3,
P(12)e4⊗e4=−e4⊗e4. (D.5)
Appendix E
q-binomial
We introduce and prove some formulas for the so called q-analysis, which are useful for the construction of quantum group, Drinfeld double and Heisenberg double ofUq(sl(2)), Drinfeld double and Heisenberg double ofUq(osp(1|2)).
Lemma 1. q-binomial formula: If A and B are elements of an algebra obeyingBA=qAB then
(A+B)n=
n
X
m=0
n m
q
AmBn−m, where
n m
q
= (n)q!
(m)q!(n−m)q!, (n)q= 1−qn 1−q .
We suppose that the q-integer (m)q is non zero for 0< m < n. By convention n0
q= 1 Proof. Assuming the result for (A+B)n−1 we have
(A+B)n−1(A+B) =
n−1
X
m=0
n−1 m
q
AmBn−1−m(A+B)
=
n−1
X
m=0
qn−1−m
n−1 m
q
Am+1Bn−1−m+
n−1
X
m=0
n−1 m
q
AmBn−m
=
n
X
m=1
qn−m
n−1 m−1
q
AmBn−m+
n−1
X
m=0
n−1 m
q
AmBn−m
=An+Bn+
n−1
X
m=1
qn−m
n−1 m−1
q
+
n−1 m
q
!
AmBn−m Then by using the identityqn−m(m)q+ (n−m)q= (n)qwe can see the proof is complete.
131
Appendix E.q-binomial 132 We present a short proof of
n k
q
= (q)n (q)k(q)n−k
, (E.1)
where, (q)n= (1−q)...(1−qn).
where we use the fact that (x;q)−1=
∞
Y
k=0
1 1−xqk =
∞
Y
k=0
∞
X
l=0
(xqk)l=
= (1 +x+x2+...)(1 +xq+x2q2+...)...(1 +xqk+x2q2k+...)...=
= 1 +x(1 +q+q2+...) +x2(1 +q+...+qk+...+q2+q3+...+q4+q5+...) +...=
= 1 + x
1−q +x2 1
1−q(1 +q2+q4+...) +...= 1 + x
1−q +x2 1 1−q
1
1−q2 +...=
= 1 +
∞
X
k=1
xk Qk
l=1(1−ql) =
∞
X
k=0
xk (q)k.
q-binomial in terms of Gb functions In the non-supersymmetric case, we have the following form of the Ramanujan summation formula
Z
dτ e−2πτ βGb(α+iτ)
Gb(Q+iτ) = Gb(α)Gb(β)
Gb(α+β) . (E.2)
From this we can get the Fourier transforms. First, by taking α→+∞ and using
Gb(x)→ζ¯b, =(x)→+∞, (E.3)
whereζb−1= ¯ζb (because ζb is a pure phase), we get Z
dτ e−2πτ β ζ¯b
Gb(Q+iτ) =Gb(β), Z
d(bt)e−2π(bt)(Q2−ir) 1
Gb(Q+ibt) =ζbGb(Q 2 −ir), b
Z
dte2πibtr e−πbtQ
Gb(Q+ibt) =ζbGb(Q
2 −ir) =g−1b (e2πbr) =e−1b (r).
Now, using the complex conjugation property of theGb
Gb(x) =e−iπ¯x(¯x−Q)Gb(¯x), (E.4) we can complex conjugate the previous expression to get
b Z
dte−2πibtr e−πbtQ
Gb(Q+ibt) = ¯ζbGb(Q 2 −ir), b
Z
dte−2πibtr e−πbtQ
Gb(Q−ibt)e−iπ(Q−ibt)(Q−Q+ibt)= ¯ζbGb(Q
2 +ir)eiπ(Q2+ir)(Q−Q2−ir), b
Z
dte2πibtr e−iπb2t2
Gb(Q+ibt) = ¯ζb 1
Gb(Q2 −ir) =gb(e2πbr) =eb(r).
Bibliography 133 Now, we can proceed with the calculation. We start from the shift property
eb(x−ib
2) =eb(x+ ib
2)(1 +e2πbx).
Using that, we defineu=e2πbq,v=e2πbp such thatuv=q2vuwithq =eiπb2. Then eb(p−q)ue−1b (p−q) =eπbq(1 +e2πb(p−q))eπbq =u+v.
Therefore,
(u+v)it=eb(p−q)uite−1b (p−q) =b2 Z
dτ1dτ2e−iπb2τ12 e2πibτ1(p−q)
Gb(Q+ibτ1)uite−πbτ2Q e2πibτ2(p−q) Gb(Q+ibτ2) =
=b2 Z
dτ1dτ2e−iπb2τ12Gb(−ibτ1)e−iπ(Q+ibτ1)(Q+ibτ1−Q)e−πbτ2Qe2πibτ1(p−q)uite2πibτ2(p−q) Gb(Q+ibτ2) =
=b2 Z
dτ1dτ2e−πb(τ2−τ1)Q Gb(−ibτ1)
Gb(Q+ibτ2)e2πibτ1(p−q)e2πibtqe2πibτ2(p−q)= (∗) and because e2πibτ1pe−2πibτ1q = e
−(2πibτ1)2
4πi e2πibτ2(p−q) = e−πib2τ12e2πibτ2(p−q) and similar expression for τ2 we have
(∗) =b2 Z
dτ1dτ2e−πb(τ2−τ1)Qe−πib2(τ1+τ2)2e2πib2tτ1 Gb(−ibτ1)
Gb(Q+ibτ2)ui(t−τ1−τ2)vi(τ1+τ2) and after the change of the integration variableτ =τ1+τ2, Therefore,
(∗) =b2 Z
dτdτ2e−2πbτ2(Q+ibt)+πbτ(Q+2ibt)−πib2τ2Gb(−ibτ+ibτ2)
Gb(Q+ibτ2) ui(t−τ)viτ Now, using the Ramanujan summation formula withα=−ibτ and β =Q+ibtwe get (∗) =b
Z
dτ eπbτ(Q+2ibt)−πib2τ2Gb(−ibτ)Gb(Q+ibt)
Gb(−ibτ+Q+ibt) ui(t−τ)viτ =
=b Z
dτ e2πib2tτ−2πib2τ2 Gb(Q+ibt)
Gb(Q+ibτ)Gb(−ibτ+Q+ibt)ui(t−τ)viτ =b Z
dτ t
τ
b
ui(t−τ)viτ, where we set
t τ
b
=b Z
dτ e2πib2τ(t−τ) Gb(Q+ibt)
Gb(Q+ibτ)Gb(−ibτ+Q+ibt).
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