q-Smidgen sl2 tangle invariants
August 12, 2016
1
Setting up the ribbon Hopf algebra
Consider the 2d-Lie algebra ii I ^ D(^)with generators w, c, over the ring Q[[]]
with bialgebra structure given by
δ(c) = 0 δ(w) = w ∧ c
Up to constants and without this is the ’standard’ structure, that reflects the
Lie algebra structure of the second Borel, if ^ is viewed as a Borel of sl2 . Hence
the name q-Smidgen sl2 . The q refers to the quantum double construction we’re
going to use to obtain a ribbon Hopf algebra.
^ has a grading, setting deg(c) = 0, deg(w) = 1. The universal enveloping
algebra A of ^ becomes a coalgebra linked to δ by ∆ − τ ◦ ∆ = δ modulo higher
order terms. A common solution is to set:
∆(c) = 1 ⊗ c + c ⊗ 1 = c1 + c2
∆(w) = ec2 w1 + w2
The counit just sends both w, c to 0 and the antipode is forced by the antipodeaxiom: S(c) = −c and µ ◦ (S ⊗ id)∆(w) = w + S(w)ec = 0 so S(w) = −we−c .
Now that we established A as a Hopf algebra, we turn to A∗ . We choose a
basis consisting of elements 1 and (cm wn )∗ for any positive integers m, n. Setting
c∗ = b and w∗ = u we now seek to express this basis in terms of monomials in
b, u. That is not automatic. Note that from the cobracket δ the elements b, u
do generate a Lie algebra T ∗ with the rule [r, s](y) = (r ⊗ s)δ(y). In our case
the relation comes out as:
[u, b] = u
Setting deg(u) = 1, deg(b) = 0 our grading extends to all of A∗ . We now want
to turn A∗ into a dually paired Hopf algebra in the following sense:
h∆(p), Q ⊗ Ri = hp, QRi
hp ⊗ q, ∆(R)i = hpq, ri
hp, 1i = (p) h1, P i = (P ) hS(p), Qi = hp, S(Q)i
It is important to check that the algebra structure on A∗ is consistent with
the coproduct on A:
= hu, wi = hub − bu, wi = hu ∧ b, ∆(w)i = hu ∧ b, ec2 w1 + w2 i = 1
We suspect that the basis {br us } is roughly dual to {cr ws } but to determine
the constants we need to compute their pairings explicitly, keeping in mind
that we want to preserve our grading. First hbr , cr i = r! by an elementary
computation using the pairing (do it!).
To determine the last pairing first note that wec = e(c+1) w = qec w. Since
it occurs often we set q = e . Also note that
∆(s−1) (w) = 1⊗1 · · ·⊗w+1⊗1 . . . 1⊗w⊗ec +1⊗1 . . . 1⊗w⊗ec ⊗ec +· · ·+w⊗ec · · ·⊗ec
In other words
∆(s−1) (w) =
s
X
j=1
wj
s
Y
eci
i=j+1
Now the pairing is
hus , ws i = hu⊗s , ∆(s−1) (w)s i = hu⊗s ,
X
e`(σ) w ⊗ ec w ⊗ e2c w ⊗ . . . e(s−1)c wi
σ∈Ss
In the last equality we only keep the terms of ∆(s−1) (w)s that have precisely one
w in every tensor factor. This means each surviving term is a product of distinct
factors and can hence be represented by a permutation. For each such term we
need to commute the ec terms to the left of the w. Each such commutation
contributes a factor q and for permutation σ it yields `(σ) such commutations,
where `(σ) is the least number of transpositions necessary to write σ. Therefore
X
hus , ws i =
q `(σ) = [b]!
σ∈Ss
s
−1
and [b]! = [1][2] . . . [s] are the Gaussian q-integers.
where [s] = qq−1
Note that when
2 = 0 we have q s = 1 + s so [s] = s + 21 s(s − 1). Also
s
1
1
[s]! = s!(1 + 2 /2) and [s]!
= s!
(1 − 2s /2) Finally
i+j
j
=
i+j
i+j
i
j
i+j
ij
(1 +
)(1 −
)(1 −
)=
(1 + )
2
2
2 2
2 2
j
2
j
In conclusion
hbi uj , ck w` i = δi,k δj,` i![j]!
i
j
b u
So the basis of A∗ dual to {ci wj } is { i![j]!
}.
The remaining Hopf algebra structure on A∗ can now be determined easily
as follows.
h∆(b), ci ⊗ cj i = hb, ci+j i = δi+j,1
so ∆(b) = b1 + b2 .
h∆(u), ci w ⊗ cj wi = hb, ci wcj wi
This leads to ∆(b) = eb2 u1 + u2 and hence also S(x) = −ue−b and S(b) = −b
and usual counit.
So far we found
2
Lemma 1. The dual A∗ has Hopf algebra structure given by
∆(c) = c1 + c2
∆(w) = w2 + ec2 w1
S(c) = −c
S(w) = −we−c
A∗ is generated by b = c∗ and u = w∗ with [u, b] = u and
∆(b) = b1 + b2
∆(u) = u2 + eb2 u1
S(b) = −b
i
S(u) = −ue−b
j
b u
Dual to the basis {ci wj } of A is the basis { i![j]!
}. Where q = e and [n] =
1.1
1−q n
1−q .
Constructing the quantum double
Next we construct the quantum double D of A using the fact that the Hopf algebra A∗ is dually paired with A. Again, as a coalgebra D = A∗ ⊗A and we require
that both A and A∗op are included in the obvious way as sub-Hopfalgebras. If
the copairing is to be a quasi-triangular structure the following algebra structure
on D is inevitable see (Etingof, Schiffman prop 12.1):
X
(P ⊗ a)(Q ⊗ b) =
hS(a0 ), Q0 iha000 , Q000 iQ00 P ⊗ a00 b
where the dashes indicate the coproduct and P, Q ∈ A∗ and a, b ∈ A. Note the
unfortunate ordering of Q00 , P that is alleviated by turning to A∗op , the dual
with opposite product.
In D the element k = k ⊗ 1 is not central. The new commutation relations
are:
[w, c] = w
[u, b] = −u [c, b] = 0
[w, b] = w
[c, u] = u [w, u] = (q−1)uw+1−eb ec
(note the extra minus sign because we are in A∗op !) As an illustration of the use
of the general product formula we derive the last commutation relation from it.
First write
∆(2) (u) = u3 + u2 eb3 + u1 eb2 eb3
(S ⊗ id ⊗ id)∆(2) (w) = w3 + w2 ec3 − w1 e−c1 ec2 ec3
Taking P, b = 1 and a = w, Q = u in the product formula now yields:
wu = 1 + quw − hwe−c , uiheb , ec i = 1 + quw − eb ec
Note that in hwe−c , ui = hq −1 e−c w, ui = q −1 it is essential that we first
commute the c to the left of w before evaluating the pairing by construction of
our dual basis. Recall that we−c = e−(c+1) w = q −1 e−c w. This is a delicate
yet crucial point.
uj bi
The basis {ci wj } of A is dual to the basis { i![j]!
} of A∗op . Hence the universal
R-matrix now comes out as
R=
X uj bi ⊗ ci wj
i![j]!
R−1 = (id ⊗ S −1 )R =
i,j≥0
X
(−1)i+j
i,j≥0
3
uj bi ⊗ (we−c )j ci
i![j]!
For the inverse we used S −1 (w) = −ec w and S −1 (c) = −c. Note in passing
that S 2 6= id, for example S 2 (w) = q −1 w.
The element b̄ = b − c is central since [b, u] = [c, u] and [b, w] = [c, w] and
[b, c] = 0. We view b̄ as more fundamental and will name it b in the q-smidgen
sl2 Hopf algebra:
Definition 1. The q-smidgen sl2 Hopf algebra smq sl2 is the Hopf algebra D
where we replace b by b+c and take the new b to be a central element (previously
called b̄). Setting [x, y]q = xy − qyx, the relations are:
[b, x] = 0
1.2
[w, c] = w
[c, u] = u
[w, u]q = 1 − eb e2c
commutation relations in smq sl2
In this section we collect useful commutation relations in smq sl2 . Set t = eb a
central element that occurs all the time.
Lemma 2. Set q = e and T = e2c and [w, u]q = r = 1 − t − 2tc = 1 − tT
1. uj ci = (c − j)i uj
uT = q −2 T u
2. wj ci = (c + j)i wj
wT = q 2 T w
3.
k `
w u =
X
q
(k−j)(`−j)
j
k
j
`
j
[j]!rj u`−j wk−j
Proof. The first five items are easy to prove so we focus on the final formula.
We first prove the case s = 1 by induction on k:
wk u = q k uwk + [k]rwk−1
The induction basis is the original commutation relation in the quantum double
wu = quw + r. If we recall [k + 1] = [k] + q k then the induction step is not hard:
wk+1 u = wq k uwk + w[k]rwk−1 = q k+1 uwk+1 + rq k wk + [k]rwk
= q k+1 uwk+1 + [k + 1]rwk
To finish the proof we do another induction to ` and we already covered the
base case ` = 1. For the induction step:
X
k
`
wk u`+1 =
q (k−j)(`−j)
[j]!rj u`−j wk−j u =
j
j
j
X
q
(k−j)(`−j)
j
X
j
k
j
`
j
q (k−j)(`+1−j)
[j]!rj u`−j (q k−j uwk−j + [k − j]rwk−j−1 ) =
k
j
4
`
j
[j]!rj u`+1−j wk−j +
X
q (k−j)(`−j)
j
X
k
j+1
q (k−j)(`+1−j)
j
X
q (k−j+1)(`+1−j)
j
Since
`+1
j
k `+1
w u
=
k
j
k
j
`
j
[j + 1]!rj+1 u`−j wk−j−1 =
[j]!rj u`+1−j wk−j +
`
j−1
`
j−1
=
X
(k−j)(`+1−j)
q
`
j
+q
`
j
`+1−j
j
k
j
[j]!rj u`+1−j wk−j =
this simplifies to
`+1
j
[j]!rj u`+1−j wk−j
as advertised.
In what follows we assume 2 = 0 and α, β and other Greek letters stand for
rational functions in b and possibly other commuting variables.
Also we define the q-exponential to be the series
ezq =
∞
X
zn
[n]!
n=0
Lemma 3. Set s = 1 − t and [w, u]q = r = s − 2tc = 1 − tT and Muw (γ) =
P k uk w k
P k w k uk
kγ
kγ
[k]! and Mwu =
[k]! .
1.
αc
eβu
= eαc eeq
q e
−α
βu
αc
eβw
= eαc eeq
q e
α
βw
2.
βu
rαβ
αw
αβuw αw
eαw
(1 + αβ∂β ∂α )eβu
= erαβ
eβu
eq
q eq = eq
q eq
q
q eq
3. Set µ =
γ
1−γs
µ
Mwu (γ) =
γ
0
and denote by Muw
(γ) the derivative with respect to γ.
µ
µs
0
00
(1 − 2µtc)Muw (µ) + ( − 2µtc)Muw (µ) + (
+ 1)Muw (µ)
γ
2
4. Set ν = 1 + γs, then
1
−1
eαw
(ν+∂α ∂γ −2tc∂α )Muw (γ)eανw
+ γsν −1 (1+γsν −1 )∂α2 Muw (γ)eανw
q Muw (γ) = ν
q
q
2
5.
−1
Muw (γ)eβu
+2tγν −1 (1−c+∂γ −
q = (s
5
s
+γsν −1 )+∂γ ∂β + γs(ν+γ)∂β2 )eβνu
Muw (γ)
q
2t
2
Proof. For part one we write out the definition of the q-exponential and use
un eαc = un eα(c−n)
X βn
X βn
X βn
−α
αc
eβu
un eαc =
eα(c−n) un = eαc
(e−α )n un = eαc eeq βu
=
q e
[n]!
[n]!
[n]!
n
n
n
And likewise wn eαc = wn eα(c+n) , so
X βn
X βn
X βn
α
n αc
α(c+n) n
αc
αc
w
e
=
e
w
=
e
(eα )n wn = eαc eeq βw
eβw
e
=
q
[n]!
[n]!
[n]!
n
n
n
For part two note that q-exponentials become exponentials when there is an
in front and it makes no matter whether we take the usual or the q-derivative:
X gz g
X gz g
=
= zezq = ∂α eαz
q
[g]!
g!
g
g
The left hand side can be written as:
X αm β n
m
n
(m−j)(n−j)
αw βu
q
[j]!rj un−j wm−j =
eq eq =
j
j
[m]![n]!
m,n,j
Adopting g = m − j and h = n − j and j as our new summation variables this
sum simplifies to
βu
eαw
q eq =
X αg+j β h+j
X αg β h
rj h g
q gh
u w = erαβ
q gh uh wg
q
[g]![h]!
[j]!
[g]![h]!
g,h,j
g,h
Using 2 = 0 we may set q gh = 1 + gh so:
βu
eαw
q eq =
X
(1 + gh)
g,h
X h(βu)h g(αw)g
αg β h h g
αw
αβr
u w = eαβr
eβu
q
q eq + eq
[g]![h]!
[g]![h]!
g,h
αw
αβr
αw
αw
= eαβr
eβu
∂β ∂α eβu
= eαβr
(1 + ∂β ∂α )eβu
q
q eq + eq
q eq
q
q eq
The proof of the third identity is very similar:
Mwu (γ) =
X γ k uk wk
k
[k]!
2
X γk
k
(k−j)2
=
q
[j]!rj uk−j wk−j
j
[k]!
k,j
Now pass to a sum over j and i = k − j to get:


X γ i+j 2 [i + j]! i + j X γi 2 X i + j =
qi
[j]!rj ui wi =
qi 
(γr)j  ui wi
j
j
[i
+
j]!
[i]![j]!
[i]!
i,j
i
j
The sum over
Qn−1j can be carried out explicitly in terms of the Pochhammer symbol
(z; q)n = f =0 (1 − zq f ) as follows.
X i + j 1
zj =
j
(z;
q)
i+1
j
6
In our case 2 = 0 and z = rγ implies
1 − zq f = 1 − γ(s − 2tc)(1 + f ) = 1 − γs − γ(sf − 2tc) =
γ
(1 − µ(sf − 2tc))
µ
and so
i
1
µi+1 Y
µ
s
= i+1
1+µ(sf −2tc) = ( )i+1 (1+µ(i(i−1) +i(s−2tc)−2tc))
(γr; q)i+1
γ
γ
2
f =0
2
With q i = 1 + i(i − 1) + i we found
Mwu (γ) =
µ
γ
X γi µ
µs
( )i+1 (1 + (i(i − 1)( + 1) + i(µ(s − 2tc) + 1) − 2µtc))ui wi =
[i]! γ
2
i
µ
µs
0
00
(1 − 2µtc)Muw (µ) + ( − 2µtc)Muw
(µ) + (
+ 1)Muw
(µ)
γ
2
since µs + 1 = µγ .
Next up is part 4.
eαw Muw (γ) =
X αm γ k
X αm γ k
k
m
w m uk w k =
q (k−j)(m−j)
[j]!rj uk−j wm+k−j
j
j
[k]![m]!
[k]![m]!
k,m,j
k,m
Now pass to a sum over j and i = k − j to get:
X αm γ i
X m im
=
q
(γrq −i )j ui wi+m
j
[m]![i]!
j
k,i
Recall that
=
m
j
=
m
j
(1 + 2 j(m − j)) so
X m
X αm γ i
j(m − 1) − j(j − 1)
(1 + q im
)(γrq −i )j ui wm+i
j
[i]![m]!
2
j
k,i
=
X αm γ i
q im (1 + γrq −i )m ui wi+m +
[i]![m]!
k,i
+γs
X αm γ i m(m − 1)
X αm γ i m(m − 1)
(1+γs)m−1 ui wi+m −(γs)2
(1+γs)m−2 ui wi+m
[i]![m]!
2
[i]![m]!
2
k,i
k,i
Set ν = 1 + γs, and note q im (1 + γrq −i )m = ν m−1 (ν + mi − 2mγtc) so that
eαw Muw (γ) = ν −1
X (αν)m γ i
k,i
[i]![m]!
7
(ν + mi − 2mγtc)ui wi+m +
+γsν −1
X (αν)m γ i m(m − 1)
[i]![m]!
k,i
2
ui wi+m −(γsν −1 )2
X (αν)m γ i m(m − 1)
k,i
[i]![m]!
2
ui wi+m
1
= ν −1 (ν + ∂α ∂γ − 2tc∂α )Muw (γ)eανw
+ γsν −1 (1 + γsν −1 )∂α2 Muw (γ)eανw
q
q
2
For part five we argue as in part four:
X γmβk
X γmβk
m
k
m m k
m (m−j)(k−j)
u
u
Muw (γ)eβu
=
w
u
=
q
[j]!rj uk−j wm−j =
q
j
j
[m]![k]!
[m]![k]!
m,k
m,k
j
Set rk = r − 2tk so um rj = r−m
um and if we set m − j = i we find
=
X k X γiβk
q ik
(γr−i−j q −i )j uk+i wi
j
[i]![k]!
j
i,k
=
X k X γiβk
q ik
(1 + j(k − j))(γr−i−j q −i )j uk+i wi
j
[i]![k]!
2
j
i,k
X γiβk
X k ik
=
q
γ j (r−i−j q −i )j uk+i wi
[i]![k]!
j
j
i,k
X γiβk X k
+
(j(k − 1) − j(j − 1))(γs)j uk+i wi
2
[i]![k]! j
j
i,k
−i
Now r−i−j q = (s − 2tc + 2(i + j)t)(1 − i) = s − 2tc + 2jt + i(2t − s) so
(r−i−j q −i )j = sj−1 (s + j(−2tc + 2(j − 1)t + 2t + i(2t − s)) therefore, recalling
ν = 1 + γs
= s−1
X γiβk
s
q ik (ν k + γ2t(1 − c + i − )ν k−1 + 2tγ 2 sν k−2 )uk+i wi
[i]![k]!
2t
i,k
+
X γiβk
k(k − 1)γs(1 + 2γs)(1 + γs)k−2 )uk+i wi
2
[i]![k]!
i,k
Therefore
−1
Muw (γ)eβu
+2tγν −1 (1−c+∂γ −
q = (s
s
+γsν −1 )+∂γ ∂β + γs(ν+γ)∂β2 )eβνu
Muw (γ)
q
2t
2
8