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I was wondering, if the presentation of Lie algebras in terms of (co)differantial graded coalgebras can be carried over to the setting of Lie Rineard pairs.
To be more precise:
If $\mathfrak{g}$ is a Lie algebra seen as $\mathbb{Z}$-graded, but concentrated in degree one only (i.e using tensor grading), then its reduced(!) exterior power $\wedge\mathfrak{g}$ can be seen as a locally nilpotent graded symmetric coalgebra, with the shuffle coproduct and the Lie bracket can be enoded into a (co)differential on $\wedge\mathfrak{g}$. Thats a common structure and a way to encode $L_\infty$-algebras as well.
However if $A$ is a commutative, associative algebra with unit and $(A,\mathfrak{g})$ is a Lie Rinehard pair, such that the action of $\mathfrak{g}$ on $A$ is not trivial, then on a first sight this doesn’t work anymore, since the (co)differential of the coalgebra $\wedge\mathfrak{g}$ does not behave well with respect to the $A$-module structure of $\wedge\mathfrak{g}$, since in general
$d(a\cdot (x_1\wedge x_2))\neq d((a\cdot x_1)\wedge x_2)\neq d(x_1\wedge(a\cdot x_2))$
for ’scalars’ $a\in A$ and vectors $x_1,x_2\in\mathfrak{g}$. Rinehard then defined another structure that has a well defined codifferential. This structure is $U(g)\otimes \wedge\mathfrak{g}$, where $U(\mathfrak{g}$ is the univer. envel. algebra of $\mathfrak{g}$ and in case of the Lie Rinehard pair of vector fields and functions, its dual is the usual DeRahm complex.
However, does anyone know, if there is a (co)differential on the $A$-module $\wedge\mathfrak{g}$ itself?
Maybe the coalgebraic formalism has to be extended to not necessarily locally nilpotent cofree coalgebras, since we now have a non trivial ’degree zero’ part and the REDUCED symmetric coalgebra as in the case of $L_\infty$-algebras isn’t sufficient anymore.
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