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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.


After thinking I realize there is the bar construction $B(A,A,L)$. Is there anything smaller in this special case?

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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 $[[x,y],[z,w]]=-[[z,w],[x,y]]$.

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I think you get a relation from $[X_1,X_2]=-[X_2,X_1]$ so that $X_1\cdot e_2=-X_2\cdot e_1$. – Don Stanley Mar 1 '10 at 14:07
Thanks this seems to be helpful and close to what I actually need. – Don Stanley Mar 1 '10 at 19:45
Silly me! I tried to get a relation from the Jacobi identity but it turned out to be a relation in the enveloping algebra, didn't occur to me to use anti-symmetry... In any case, the generating series $p$ for the basis is a symmetric function and this shows that $\sigma_2$ appears in it. – Torsten Ekedahl Mar 2 '10 at 5:15

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.

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This is great. Thanks a lot. I need to think about it for a bit. – Don Stanley Mar 1 '10 at 13:39
For the differential I mean that $T(V)$ is a differential graded algebra and $L(V)$ is a differential graded module over it. Then it would be great to have $L(V)$ as the cone on a map in the category of $T(V)$ modules (as in your resolution), but I guess that's not possible. – Don Stanley Mar 1 '10 at 13:53
Slightly less good would be to have a $T(V)$ dg-module quasi-isomorphic to $L(V)$ that was free as a $T(V)$ module (like the bar construction). – Don Stanley Mar 1 '10 at 13:58
You can take the total complex of the resolution I constructed above: that is a free $T(V)$-module with a differential. The map induced by my map $T(L)\otimes L(V)\to L(V)$ is then a quasi-iso. – Mariano Suárez-Alvarez Mar 1 '10 at 14:05
What is the differential on the $T(V)\otimes V\otimes T(V)$ (or $T(V)\otimes V\otimes L(V)$) term? – Don Stanley Mar 1 '10 at 14:13

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