Algebra and Applications 2. Группа авторов
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[1.14]
We say that the filtered bialgebra ℋ is connected if ℋ0 is one-dimensional. There is an analogue of Proposition 1.6 in the connected filtered case, the proof of which is very similar:
PROPOSITION 1.7.– For any x ∈ ℋn, n ≥ 1, we can write:
[1.15]
The map
As an easy corollary, the degree of the antipode is also zero in the connected case, that is, S(ℋn) ⊆ ℋn for any n. This is an immediate consequence of the recursive formulae [1.19] and [1.20] below.
Any graded bialgebra, or the Hopf algebra, is obviously filtered by the canonical filtration associated with the grading:
[1.16]
and in that case, if x is a homogeneous element, x is of degree n if and only if |x| = n.
1.3.4. The convolution product
An important result is that any connected filtered bialgebra is indeed a filtered Hopf algebra, in the sense that the antipode comes for free. We give a proof of this fact as well as a recursive formula for the antipode using the convolution product: let ℋ be a (connected filtered) bialgebra, and let
PROPOSITION 1.8.– The map
PROOF.– The first statement is straightforward. To prove the second, let us consider the formal series:
Using (e — φ)(1) = 0, we have (e — φ)*k(1) = 0 immediately, and for any x ∈ Ker ε:
[1.17]
When x ∈ ℋn, this expression vanishes and then for k ≥ n + 1. The formal series then ends up with a finite number of terms for any x, which proves the result. □
COROLLARY 1.1.– Any connected filtered bialgebra ℋ is a filtered Hopf algebra. The antipode is defined by:
[1.18]
It is given by S(1) = 1 and recursively by any of the two formulae for x ∈ Ker ε:
PROOF.– The antipode, when it exists, is the inverse of the identity for the convolution product on
fulfilled by any x ∈ Kerε.
Let
From now on, we will suppose that the ground field k is of characteristic zero. For any x ∈ ℋn, the exponential:
[1.21]
is a finite sum (ending up at k = n). It is a bijection from
[1.22]
This sum again ends up at k = n for any x ∈ ℋn. Let us introduce a decreasing filtration on
[1.23]
Clearly,