Resolution of a free lie algebra as a module over its universal enveloping algebra. Let $L=L(V)$ be a free Lie algebra on a vector space $V$ and $A=T(V)$ the tensor algebra. Make $L$ into a module over $A$ consistent with the formula $a\cdot \alpha=[a,\alpha]$ for $a\in V$ and $\alpha\in L$. 
What is a canonical resolution of $L$ by free $A$ modules? I'm really most interested in the case where there is a grading and a differential. 
Edited:
After thinking I realize there is the bar construction $B(A,A,L)$. Is there anything smaller in this special case?
 A: There is a very small resolution. Everything is graded so we can in fact speak
of minimal resolutions. The most refined version is to make the basis (which I
for simplicity assume is finite of cardinality $n$) is graded by itself so that
everything becomes $\mathbb N^n$-graded. (Assume that the generators are
$X_1,\dots,X_n$.) Then the $X_i$ form a minimal set of generators of $L$ as a
$T$-module and the first step of a minimal resolution has the form
$\bigoplus_iT[-e_i]\to L \to 0$ (where $e_i$ is the standard basis of $\mathbb
N^n$). Now $T$ has global dimension $1$ so that the kernel $K$ is projective and
being graded is free graded. A quick look doesn't reveal any elements in the
kernel but in principle we can tell in which degrees the generators are using
multi-degree Hilbert series:
We have that the Hilbert series $H_T$ of $T$ is $1/(1-\sigma_1)$, where
$\sigma_1=\sum_ix_i$ and if the Hilbert series $L_T$ of $L$ is $\sum_{\alpha}a_\alpha x^\alpha$, then we
have $1/(1-\sigma_1)=\prod_\alpha(1-x^\alpha)^{a_\alpha}$ which determines the coefficients
$a_\alpha$ (done explicitly using Möbius inversion, see for instance Bourbaki, Lie
algebras, Chap. II). If $p$ is the generating series for the basis of $K$ we get
$\sigma_1H_T=pH_T+L_T$ which determines $p$.
In any case this only gives information on a possible resolution (even though my guess is people have
done it otherwise you could perhaps guess what happens using the above to see where the first relations
would appear). 
[Added later]
As penance for failing to find the first very simple relation I computed some terms of the generating series
of the basis. Notice that it is clearly a symmetric function in the $x_i$ and in fact we have a natural
$GL(V)$-action on the span of a basis so that this symmetric function is in fact a character of $GL(V)4. We 
have the following:
In degree $2$ we have $\sigma_1^2-\sigma_2$ (not $\sigma_2$ as I claimed in a comment). This is the character of $S^2V$ and in fact the relations $X_ie_j+X_je_i$ account for that part.
In degree $3$ we have $\sigma_3$ which is the character of $\Lambda^3V$.
In degree $4$ we have $\sigma_2^2-\sigma_1 \sigma_3$.
I have put the Mathematica notebook with the calculations here.
[Added later still]
I did some calculations of the representations involved. Using the parametrisation
of irreducible representations in terms of partitions I get the following (to
possibly avoid confusion as to whether I use a partition or its dual, $(n)$ is
$S^nV$ and $(1^n)$ is $\Lambda^nV$):
2: $(2)$
3: $(1^3)$
4: $(2^2)$
5: $(3, 1^2)$
6: $(5,1) + 2(4,1^2) +  3(3,2,1) + (2^3) + 2(3,1^3) + (2^21^2) + (2,1^4)$
Not a very discernible pattern. (I don't think there is any reason why there
should be one.) A different way (and in some sense more concrete) of thinking
about the problem is to use polynomial functors. Hence we have that the map
$T(V)\bigotimes V \to L(V)$ is functorial in $V$. McDonald's theory of polynomial
functors (the consequence I'll be using can easily be proved directly) gives
that everything is determined by what the map does on monomials where all the
variables are distinct. Rather than going through the details of this let me
give the concrete examples:
In degree $2$ we have the element $x\cdot y+y\cdot x$ in the kernel (I use the dot to
distinguish the second factor in $T(V)\bigotimes V$) as $x\cdot y$ maps to $[x,y]$
and $y\cdot x$ to $[y,x]=-[x,y]$. This implies that the kernel in general is spanned
by tensors of the form $u\cdot v+v\cdot u$ for $u,v\in V$. Furthermore, the action of the
symmetric group $\Sigma_2$ on $x\cdot y+y\cdot x$ is trivial and then spans the trivial
representation which under the Weyl-McDonald equivalence corresponds to $S^2V$
so that the kernel has that form.
In degree $3$ we have the elements coming from degree $2$ which are
$z(x\cdot y+y\cdot x)$, $y(x\cdot z+z\cdot x)$ and $x(y\cdot z+z\cdot y)$. We also have $xy\cdot z+zx\cdot y+yz\cdot x$
coming from the Jacobi identity. To see that they span the whole kernel it is
convenient to compute in $L(V)$ using a Hall basis (see Bourbaki again). With
the appropriate choice of ordering on monomials (which is part of the building
up of a Hall basis) we have a Hall basis for the monomials for which each variable
occurs only once which is $[z,[x,y]]$ and $[y,[x,z]]$ and then the different
monomials map as follows to linear combinations of Hall monomials:
$xy\cdot z \mapsto [ y,[ x,z] ] -[ z,[ x,y]] $
$xz\cdot y \mapsto [ z,[ x,y] ] -[y,[ x,z] ]$
$ yx\cdot z \mapsto [ y,[ x,z] ]$
$ yz\cdot x \mapsto -[ y,[ x,z] ]$
$ zx\cdot y \mapsto [ z,[ x,y] ]$
$ zy\cdot x \mapsto -[ z,[ x,y] ] $
This shows that the kernel is indeed spanned by the relations we have
found. Furthermore, we see that $xy\cdot z+zx\cdot y+yz\cdot x$ modulo the others is
antisymmetric showing that the new basis elements give a $\Lambda^3V$ (as the
signature representations correspond to the exterior power). Note that the space
spanned by $xy\cdot z+zx\cdot y+yz\cdot x$ is not $\Sigma_3$-invariant so does not give an
embedding of $\Lambda^3V$ in the kernel. However, anti-symmetrising it gives
$xy\cdot z+zx\cdot y+yz\cdot x-yx\cdot z-zy\cdot x-xz\cdot y$ and hence the space spanned by
$uv\cdot w+wu\cdot v+vw\cdot u-vu\cdot w-wv\cdot u-uw\cdot v$ for $u,v,w\in V$ spans a $\Lambda^3V$.
This can be continued but I have not done so. Note, from the point of view of
operads the Lie operad is the quotient by a free operad using the anti-symmetry
and Jacobi relations. This may make it somewhat confusing as to why we get more
basis elements in degrees higher than $3$. The reason is that from the current
point of view we allow ourselves only to generate new relations (which are not
part of the basis) by commuting with arbitrary elements whereas in the operadic
description we also allow the substitution arbitrary monomials in old relations:
For instance, we get from $[x,y]=-[y,x]$ that $[[x,y],[z,w]]=-[[z,w],[x,y]]$.
A: There is a very short resolution of $T(V)$ as a $T(V)$-bimodule, $$0\to T(V)\otimes V\otimes T(V)\to T(V)\otimes T(V)\twoheadrightarrow T(V)$$ with the first map being the unique one maps of $T(V)$-bimodules such that $1\otimes v\otimes 1\in T(V)\otimes V\otimes T(V)$ maps to $v\otimes 1-1\otimes v\in T(V)\otimes T(V)$, and the second one simply by the product on $T(V)$. Now the Lie algebra $L(V)$ is a $T(V)$-module (on the left, say), so we can apply the functor $(\mathord-)\otimes_{T(V)}L(V)$ to out complex, getting, up to standard identifications, $$0\to T(V)\otimes V\otimes L(V)\to T(V)\otimes L(V)\twoheadrightarrow L(V),$$ with induced maps. This is a $T(V)$-projective resolution of $L(V)$.
This is a graded resolution, if you want to consider the natural grading on $T(V)$, $L(V)$ and the induced gradings on the complex. `Handling a differential' depends on what you mean by that.
