Now we can write h = tqla with a, l ∈ N. If q 6= 1 we may assume that q
does not divide a. We want to show that a = 1. So assume a > 1. In this case we can construct a k-linear map φ2 :kX⊗kX→k with
∀(b, b0)∈(B×B)\ {(vtql, vtql(a−1))}:φ2([b]⊗[b0]) = 0,
φ2([v]tq
l
⊗[v]tql(a−1)) = 1, φ2(I⊗kX+kX⊗I) = 0.
Using that γv,v is a primitivet-th root of unity (resp. γv,v = 1 andt = 1) we
obtain that in k 0 = φ2∆(T) = tqla tql γv,v = qla ql =a 1 =a.
This is a contradiction to the assumptions we made on a. Thus h = tql. In particular if chark = 0, then q = 1 and because t = h > 1 we obtain γv,v 6= 1.
2.4
Transfer to right triangular braidings
In principle one could do a similar proof as above for right triangular braid- ings, but an easy argument shows that the right triangular case follows from the left triangular case. Obviously c is a right triangular braiding if and only if τ cτ is left triangular, where τ denotes the usual flip map. The key observation is
Proposition 2.4.1. Let (R, µ, η,∆, ε, c) be a braided bialgebra. Then also Rop,cop := (R, µτ, η, τ∆, ε, τ cτ) is a braided bialgebra.
Proof. Let µop := µτ and ∆cop := τ∆. Of course (R, µop, η) is an algebra,
(R,∆cop, ε) is a coalgebra and (R, τ cτ) is a braided vector space. Checking the compatibility ofµop, η,∆cop, εwith τ cτ is tedious. We will do one exam- ple, namely the calculation thatτ cτ◦(µop⊗R) = (R⊗µop)(τ cτ⊗R)(R⊗τ cτ).
We calculate (R⊗µop)(τ cτ ⊗R)(R⊗τ cτ) = (R⊗µ)(R⊗τ)(τ ⊗R)(R⊗τ)(R⊗τ)(cτ ⊗R)(R⊗τ cτ) = (R⊗µ)(τ⊗R)(R⊗τ)◦(τ ⊗R)(R⊗τ)(c⊗R)(τ ⊗R)(R⊗τ cτ) = τ(µ⊗R)◦(R⊗c)(τ⊗R)(R⊗τ)(τ ⊗R)(R⊗τ cτ) = τ(µ⊗R)(R⊗c)(R⊗τ)(τ⊗R)(R⊗c)(R⊗τ) = τ(µ⊗R)(R⊗c)(c⊗R)◦(R⊗τ)(τ ⊗R)(R⊗τ) = τ c(R⊗µ)◦(τ⊗R)(R⊗τ)(τ ⊗R) = τ cτ(µ⊗R)(τ ⊗R) = τ cτ(µop⊗R),
where we use (in this order): τ2 = id
V⊗V, the braid equation for τ, µ, c
commute withτ, again the braid equation forτ and τ2 = id
V⊗V,ccommutes
with τ, µ commutes with c and the braid equation for τ and finally again thatµcommutes withτ. The other calculations work similarly (use graphical calculus as a tool for intuition).
Finally we have to check that ∆ : R → R⊗R and ε : R → k are algebra morphisms, whereR⊗Ris an algebra with multiplication (µ⊗µ)(R⊗τ cτ⊗R). Forε this is trivial. For ∆ we have to check
∆copµop= (µop⊗µop)(R⊗τ cτ ⊗R)(∆cop⊗∆cop). As R is a braided bialgebra the left hand side is
τ∆µτ =τ(µ⊗µ)(R⊗c⊗R)(∆⊗∆)τ. Now because ∆, µ commute withτ this is equal to
(µ⊗µ)(R⊗τ ⊗R)(τ ⊗τ)(R⊗τ cτ ⊗R)(τ ⊗τ)(R⊗τ⊗R)(∆⊗∆). Thus it suffices to show
(R⊗τ⊗R)(τ⊗τ)(R⊗τ cτ⊗R)(τ⊗τ)(R⊗τ⊗R) = (τ⊗τ)(R⊗τ cτ⊗R)(τ⊗τ), but this is trivial (check on elements).
Related material can be found in [2].
Assume now that (V, c) is a braided vector space. Denote the braided tensor bialgebra defined in Section 1.4 by (T(V, c), µ, η,∆c, ε, c). As an algebra this
2.4. Right triangular braidings 59
Proposition 2.4.2. Let (V, c) be a braided vector space. Let
φ :T(V, c)→T(V, τ cτ)op,cop
be the unique algebra morphismT(V)→T(V)op given by φ|V = id
V. Then
φ is an isomorphism of braided bialgebras. Forv1, . . . , vn∈V we have
φ(v1⊗. . .⊗vn) = vn⊗. . .⊗v1.
Proof. Lemma 1.4.15 gives us the existence of φ as a morphism of braided bialgebras because (V, c) is a braided subspace of T(V, τ cτ)op,cop. By con- struction we see that the map φ has the form given in the proposition and that it is bijective.
Now we can prove the existence of the PBW basis in the right triangular case.
Theorem 2.4.3. Assume that (V, c) is a finite-dimensional right triangular
braided vector space and I ( T(V, c) is a braided biideal. Then there is a totally ordered subset S ⊂ T(V, c) and a height function h :S → N∪ {∞} such that the images of the PBW set generated by S and h form a basis of T(V, c)/I.
Let
φ :T(V, c)→T(V, τ cτ)op,cop be the isomorphism from Proposition 2.4.2. We have
S =φ−1(Sφ(I)) and h=hφ(I)φ,
and the order on the set S is the opposite of the order on Sφ(I).
Proof. Asφ(I) is a braided biideal inT(V, τ cτ)op,copit is also a braided biideal
inT(V, τ cτ). As c is right triangular we have that τ cτ is left triangular. So we find a setSφ(I) ⊂T(V, τ cτ) with a total ordering <and a height function
hφ(I) : S → N∪ {∞} such that the PBW set generated by these data in
T(V, τ cτ) is a basis for a complement of φ(I). The PBW set generated in T(V, τ cτ)op,cop by Sφ(I) with reversed order and height function hφ(I) is the
same set and thus also a basis for a complement of φ(I). The claim follows by transferring this set toT(V, c) viaφ−1.
Corollary 2.4.4. Let (V, c) be a braided vector space. Then
B(V, τ cτ)'B(V, c)op,cop as braided graded Hopf algebras.
Proof. Denote the braiding of the braided bialgebraB(V, c) by ˆc. By Propo- sition 2.4.1B(V, c)op,cop is a braided bialgebra with braiding τcτˆ . It is easy to
check that B(V, c)op,cop has the properties of the Nichols algebra of (V, τ cτ)
from Definition 1.4.7.
2.5
Application to pointed Hopf algebras with
abelian coradical
In this section we will show how to obtain a PBW basis for a Hopf algebra generated by an abelian groupGand a finite-dimensionalG-module spanned by skew primitive elements. On one hand this yields a generalization of the result in [20] as there the skew primitive elements are assumed to be semi- invariants (i.e. that the group acts on them by a character). On the other hand we lose some properties of the basis as already mentioned in Remark 2.2.5.
LetA =∪n≥0An be a filtered algebra. We can define a map
π :A→grA
by setting π(0) := 0 and for all 06=a ∈A:π(a) :=a+An−1 for the unique
n ≥ 0 such that a ∈ An\An−1 (where A−1 := {0} as usual). We will use
this map to obtain PBW bases for A from homogeneous PBW bases of the associated graded algebra grA.
Proposition 2.5.1. Let A = ∪n≥0An be a filtered algebra and (P, S, <, h)
a PBW basis for grA such that P ⊂ grA(0) = A0 and S is made up of
homogeneous elements. Then there is a PBW basis (P, S0, <0, h0) of A such that for all a, b∈S0
π(a)∈S, h0(a) =h(π(a)) and a < b⇔π(a)<0 π(b).
Proof. For all s ∈ S ∩grA(n) we find ˆs ∈ An \An−1 such that π(ˆs) = s.
Define
S0 :={ˆs|s∈S}.
The map S →S0, s 7→sˆis bijective. So we can transfer the height function h and the order < toS0 obtaining h0 and <0.