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I have problems finding anything about the cofree Lie coalgebra functor $\mathcal{L}ie^c$ out there.

Basically all I found was that it appears in Harrison cohomology and that, given a $\mathbb{Z}$-graded vector space $A$, the underlaying vector space of $\mathcal{L}ie^c(A)$ is the following:

For any $p,q,n\in\mathbb{N}$, with $p+q=n$, the shuffle multiplication is

$sh_{(p,q)}:\otimes^p A \times\otimes^q A \to \otimes^n A\;;\; ((a_1\otimes\cdots\otimes a_q),(a_{q+1}\otimes\cdots\otimes a_n) )\mapsto \sum_{\sigma\in Sh(p,q)} a_{\sigma^{-1}(1)}\otimes\cdots\otimes a_{\sigma^{-1}(n)}$

were $Sh(p,q)$ is the set of all shuffle permutations. Then

$\mathcal{L}ie^c(A)_n=\bigotimes^nA /\{sh_{(p,q)}(\otimes^p A \times\otimes^q A), p+q=n, p,q,n\in\mathbb{N}\}$ is the quotient of the $n$-th tensor power by the images of all shuffle multiplications.

Now there are two questions:

1.) What is the appropriate Lie coproduct?

2.) How does 'cofreeness' works here. I.e given a linear map $f: C \to A$ from another Lie coalgebra, what is the unique extension $F: C \to \mathcal{L}ie^c(A)$?

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  • $\begingroup$ The sum should be on the RHS, not the LHS. And you need to assume $p$ and $q$ positive when you factor out the span (not the set!) of the shuffle subspaces. $\endgroup$
    – darij grinberg
    Commented Nov 16, 2014 at 18:11
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    $\begingroup$ I don't have time to leave a detailed answer but this is explained in the book "Algebraic operads" by Loday-Vallette. In the first part of the book they explain that the tensor algebra quotiented by nontrivial shuffle products is naturally dual to the space of formal Lie words. You can think of $Lie^c(A) = Lie^c(n) \otimes_{S_n} A^{\otimes n}$ where $Lie^c(n)$ is the $S_n$-module spanned by binary trees with a single root and $n$ leaves, modulo the antisymmetry and Jacobi relations. The coproduct is given by a sum over all ways of splitting trees in half by deleting an edge. $\endgroup$
    – Dan Petersen
    Commented Nov 16, 2014 at 20:49
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    $\begingroup$ The cofreeness is also clear in this picture. A tree as above looks like it describes an $n$-fold iterated comultiplication, so "apply" this tree to an element of $C$ to get an element of $A^{\otimes n}$ which is only well defined modulo the antisymmetry and Jacobi relations. $\endgroup$
    – Dan Petersen
    Commented Nov 16, 2014 at 20:50
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    $\begingroup$ Also, your cofree Lie coalgebras should really be cofree conilpotent Lie coalgebras, to be pedantic. $\endgroup$
    – Dan Petersen
    Commented Nov 16, 2014 at 20:51
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    $\begingroup$ Yes you are absolutely right, to be pedantic about the conilpotent part. Without that things are very different. $\endgroup$
    – Mark.Neuhaus
    Commented Nov 16, 2014 at 22:50

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I know about the existence of cofree Lie coalgebras from the paper by Michaelis "Lie Coalgebras", which I assume could be be the first time they appeared. You may also have a look at Griffing's "A nonisomorphism theorem for cofree Lie coalgebras". However, I believe that the explicit construction of these objects can be found in Michaelis' "Lie Coalgebras" Ph.D. dissertation or in Griffing's "Cofree Lie and Other Coalgebras" Ph.D. dissertation, but I have no access to these works so I cannot tell you more.

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