Let $G$ be a finitely presented group. To $G$ is associated in a functorial way a Malcev Lie algebra which can be constructed in several equivalent ways. Roughly speaking, it is the quotient of the completed free Lie algebra on the same number of generators as $G$, by the (formal) logarithm of the defining relations of $G$ (see edit below).


$G$ is called 1-formal if its Malcev Lie algebra is isomorphic as a filtered Lie algebra to the degree completion of a finitely presented Lie algebra with quadratic relations. (If $X$ is a path-connected topological space such that $G=\pi\_1(X)$, this is equivalent to say that the Malcev Lie algebra of $G$ is isomorphic to the holonomy Lie algebra of $X$)


It is quite well known that fundamental groups of complementary of hyperplane arrangements in $\mathbb{C}^n$ (e.g. pure braid groups) are 1-formal.


On the one hand, an important fact in the study of these groups is that they are (under some conditions on the underlying arrangement) iterated "almost-direct product" of free groups, and free groups are themselves 1-formal. An almost direct product is a semi-direct product $H\rtimes K$ for which the action of $K$ on the abelianization of $H$ is trivial. On the other hand, according to this interesting survey <http://www.arxiv.com/abs/0903.2307> the direct product of two 1-formal groups is again 1-formal, even if no proof is given (it's probably easy..)


So it is temptating to ask: 



> 
> is an almost-direct product of
>  1-formal groups again 1-formal ?
> 
> 
> 


**Edit:** Some details about the construction and its relation with almost direct product. If $G$ is an abelian group, then one can take the tensor product $G \otimes\_{\mathbb{Z}} \mathbb{Q}$ leading to a uniquely divisible abelian group. The Malcev construction extends this to any nilpotent group, leading to a uniquely divisible nilpotent group. Let $G^{(0)}=G$, $G^{(n+1)}=[G^{(n)},G]$ be the lower central serie, then the quotients $G/G^{(n)}$ are nilpotent by construction.


The Malcev completion of $G$ is the inverse limit of the $(G/G^{(n)}) \otimes\_{\mathbb{Z}} \mathbb{Q}$. It is a pro-unipotent group, hence it has a (pro-nilpotent) Lie algebra $\mathfrak g$. On the other hand, one can define a Lie algebra by 
$$gr\ G=\mathbb{Q}\otimes\_{\mathbb{Z}} \bigoplus G^{(n)}/G^{(n+1)}$$
the bracket being induced by the commutator in $G$. It as a graded Lie algebra, and in fact it is the associated graded of the filtered Lie algebra $\mathfrak g$.


A group is called 1-formal if there is an isomorphism of filtered Lie algebras $gr\ G \cong \mathfrak g$. Now, the point is that almost-direct product behave well with respect to the lower central series. If $G=G\_1\rtimes G\_2$ is an almost-direct product, then it seems to me that a result of Falk and Randell implies that we have 
$$G^{(n)}=G\_1^{(n)}\rtimes G\_2^{(n)}$$
and
$$gr\ G = gr\ G\_1 \rtimes gr\ G\_2$$.



Almost-direct product and 1-formality
Yes, the direct product of two 1-formal groups is again 1-formal, and so is the free product of two 1-formal groups. A proof is given in [arxiv:0902.1250](http://arxiv.org/abs/0902.1250), Proposition 9.2.


And no, the almost direct product of two 1-formal groups need not be 1-formal. A proof is given in the same paper, Example 8.2. The group in question is a semi-direct product of the form $G=F\_4\rtimes F\_1$, where $F\_n$ is the free group of rank $n$, which is of course 1-formal. The action of $F\_1$ on $F\_4$ is given by a certain pure braid $\beta \in P\_4$, acting via the Artin representation on $F\_4$; thus, the action is trivial on $H\_1(F\_4)$. For this extension, the ``tangent cone formula" fails: the tangent cone to the characteristic variety $V\_2(G)$ is strictly included in the resonance variety $R\_2(G)$. In view of Theorem A from the cited paper, the group $G$ is not 1-formal. 


It is worth noting that $G$ is the fundamental group of the complement of a certain link of 5 great circles in $S^3$. Alternatively, $G$ can be realized as the fundamental group of the complement of an arrangement of 5 planes in $\mathbb{R}^4$, meeting transversely at the origin (of course, this real arrangement cannot be isotoped to an arrangement of 5 complex lines in $\mathbb{C}^2$). For more details on the construction and properties of such arrangements, see [arxiv:math.GT/9712251](http://arxiv.org/abs/math.GT/9712251). In particular, the pure braid $\beta$ is described there in Propositions 4.4, 4.6, and 4.9.


