Algebra and Applications 2. Группа авторов

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Algebra and Applications 2 - Группа авторов

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that a filtered bialgebra with antipode is automatically a filtered Hopf algebra (see, for example, Montgomery (1993, Lemma 5.2.8), Andruskiewitsch and Schneider (2002) and Andruskiewitsch and Cuadra (2013)). The antipode is, however, of degree zero in the connected case: for any x ∈ ℋ, we set

      [1.14]image

      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]image

      The map image is coassociative on Ker ε and image sendsn into (ℋn-k)⊗k + 1.

      Any graded bialgebra, or the Hopf algebra, is obviously filtered by the canonical filtration associated with the grading:

      [1.16]image

      and in that case, if x is a homogeneous element, x is of degree n if and only if |x| = n.

      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 image be any k-algebra (which will be called the target algebra): the convolution product on image is given by:

image

      PROPOSITION 1.8.– The map image, given by image and e(x) = 0 for any xKer ε, is a unit for the convolution product. Moreover, the set image endowed with the convolution product is a group.

      PROOF.– The first statement is straightforward. To prove the second, let us consider the formal series:

image

      Using (eφ)(1) = 0, we have (eφ)*k(1) = 0 immediately, and for any xKer ε:

      [1.17]image

      When x ∈ ℋn, this expression vanishes and then for kn + 1. The formal series then ends up with a finite number of terms for any x, which proves the result. □

      [1.18]image

      It is given by S(1) = 1 and recursively by any of the two formulae for xKer ε:

      PROOF.– The antipode, when it exists, is the inverse of the identity for the convolution product on image. We then just need to apply Proposition 1.8 with image. The two recursive formulas come directly from the two equalities:

image

      fulfilled by any x ∈ Kerε.

      Let image be the subspace of image formed by the elements α, such that α(1) = 0. It is clearly a subalgebra of image for the convolution product. We have:

image

      From now on, we will suppose that the ground field k is of characteristic zero. For any x ∈ ℋn, the exponential:

      [1.21]image

      is a finite sum (ending up at k = n). It is a bijection from image onto image. Its inverse is given by:

      [1.22]image

      This sum again ends up at k = n for any x ∈ ℋn. Let us introduce a decreasing filtration on image:

      [1.23]image

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