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Fix a ground commutative ring $k$. A coPoisson-bialgebra is a bialgebra $H$ equipped with a linear mapping $\pi:H \rightarrow H \otimes_k H$ s.t.

  • $\pi$ is a coLie bracket
  • $\pi$ is a coderivation
  • $\pi(ab) = \Delta(a)\pi(b)+\pi(a)\Delta(b)$

I'm mostly interested in the case of cocommutative $H$. For such $H$, coPoisson structure is morally an infinitesimal deformation away from cocommutativity (the structure needed for "coquantization" preserving bialgebra structure). Also, I'm mostly interested in the Hopf algebra case but this doesn't seem important for the question

The category of left modules over $H$ is a $k$-linear Abelian category (just because H is a $k$-algebra) equipped with a tensor product functor (due to the coproduct $\Delta$ on H). For cocommutative $H$ the tensor product is symmetric. The question is:

What additional structure(s) on this category are obtained from $\pi$?

EDIT: I think I figured out the answer in case $k$ is a field. In this case the tensor product functor is exact (rather than just right exact)

Introduce $h$ a formal parameter satisfying $h^2=0$. $\pi$ defines a "deformed" coproduct on $H[h]$ given by


Thus the category of left $H[h]$-modules becomes a tensor category

Denote $\mathcal{M}$ our symmetric tensor category. The tensor product functor is $\otimes$ and its symmetric braiding is $b: X \otimes Y \rightarrow Y \otimes X$

We construct the Abelian category $\mathcal{M}[h]$. An object $X$ in $\mathcal{M}[h]$ is an object in $\mathcal{M}$ equipped with an endomorphism $h: X \rightarrow X$ s.t. $h^2=0$. A morphism in $\mathcal{M}[h]$ is a morphism in $\mathcal{M}$ which commutes with $h$

$\mathcal{M}[h]$ comes with the following functors of interest:

  • $Ker \space h: \mathcal{M}[h] \rightarrow \mathcal{M}$
  • $Coker \space h: \mathcal{M}[h] \rightarrow \mathcal{M}$
  • $i: \mathcal{M} \rightarrow \mathcal{M}[h]$ which sends $X$ to itself with $h=0$
  • $[h]: \mathcal{M} \rightarrow \mathcal{M}[h]$; Given $X \in \mathcal{M}$ we define $X[h]$ to be $X \oplus X$ equipped with the obvious action of $h$

The desired additional structure is:

  • An exact $k[h]$-linear tensor product functor $\hat{\otimes}$ on $\mathcal{M}[h]$
  • Tensor functor strucutre on $Coker \space h$

The antisymmetry of $\pi$ imposes the following additional condition:

Note that $i(X) \hat{\otimes} Y \cong X \otimes Coker \space h_Y$ which follows by observing that $h$ annihilates $i(X) \hat{\otimes} Y$ due to $k[h]$-linearity of $\hat{\otimes}$, applying $Coker \space h$ and using its tensor functor structure

We have the short exact sequence

$$0 \rightarrow X \xrightarrow{Ker \space h} X[h] \xrightarrow{Coker \space h} X \rightarrow 0$$

We apply $\hat{\otimes} Y[h]$ and use its exactness to get the short exact sequence

$$0 \rightarrow X \otimes Y \rightarrow X[h] \hat{\otimes} Y[h] \rightarrow X \otimes Y \rightarrow 0$$

This sequence yields an element $\epsilon(X, Y)$ of $Ext^1(X \otimes Y, X \otimes Y)$

Due to $b$, $\epsilon(X, Y)$ and $\epsilon(Y, X)$ belong to canonically isomorphic spaces. We demand

$$\epsilon(X, Y)+\epsilon(Y, X)=0$$

The problem is I need the case in which $\otimes$ is merely right exact. What currently confuses me is that in the bialgebra picture the last sequence is still exact for general $k$ but I can't prove it in the category-theoretic language

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I may be a little naive here but since any bialgebra is coPoisson with respect to the null coderivation is there really something to be expected on the category of all H-modules? – Nicola Ciccoli Feb 13 '12 at 10:38
@Nicola, the structure depends on the choice of coPoisson bracket, not merely on its existence. See also my recent edit – Squark Feb 13 '12 at 21:17

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