# Different definitions of formality for groups

Let $$X$$ be a space with fundamental group $$G$$. Recall that the de Rham fundamental group of $$X$$ is the inverse limit of the Malcev completions of the nilpotent truncations of $$G$$. This has a Lie algebra, which I will denote by $$\mathfrak{g}$$. The Lie algebra $$\mathfrak{g}$$ has a natural filtration, and thus has an associated graded Lie algebra $$\text{gr}\ \mathfrak{g}$$. In the literature, I have seen two different definitions of what it means for $$G$$ to be formal:

1. The Lie algbera $$\mathfrak{g}$$ is isomorphic to its associated graded $$\text{gr}\ \mathfrak{g}$$.

2. The Lie algebra $$\mathfrak{g}$$ has a presentation with only quadratic relations.

There is also a notion of $$X$$ being formal:

1. The minimal model of $$X$$ is quasi-isomorphic to the cohomology algebra (with the trivial differential).

I assume that 3 implies 1 and 2 (though presumably it is stronger).

Question: What is the precise relationship between the above notions? Is there a good place to read proofs of whatever implications exist?

• Clearly 2 implies 1, and the converse doesn't hold (for instance the 3-dimensional Heisenberg Lie algebra satisfies 1 and not 2).
– YCor
Commented Jan 16, 2020 at 22:00
• In addition to Denis excellent answer you might want to have a look at the very nice survey by Suciu-Wang in arxiv.org/abs/1504.08294 Commented Jan 17, 2020 at 9:21

## 1 Answer

There is a notion of $$q$$-equivalence between cdgas: it is a chain of morphisms each being isomorphism on cohomology in degrees $$\leq q$$ and monomorphism in degree $$q+1$$. Now, we call something $$q$$-formal if it is $$q$$-equivalent to its cohomology. Obviously, a space $$X$$ is $$1$$-formal if and only if $$K(\pi_1(X), 1)$$ is $$1$$-formal because killing homotopy groups map $$X \to K(\pi_1(X), 1)$$ is evidently an 1-equivalence.

Define holonomy algebra $$hol(X)$$ of a cdga as quotient of free Lie algebra on $$(H^1)^*$$ by ideal spanned by image of comultiplication $$\mu^*: (H^2)^* \to (H^1)^* \wedge (H^1)^*$$. It's obviously a graded quadratic Lie algebra.

Notice that 1-equivalences give isomorphisms on holonomy Lie algebras. So, for spaces, it depends only on fundamental group, because if you have $$G \cong \pi_1(X) \cong \pi_1(Y)$$, then both map into $$K(G, 1)$$ with maps being $$1$$-equivalences.

Malcev completion of a group $$\mathfrak m (G)$$ is defined as Lie algebra of primitive elements in $$\lim_{\leftarrow} \Bbb QG/IG^n$$ where $$IG$$ is augmentation ideal. There is a homomorphism from Magnus algebra of a group $$L(G) := \bigoplus \gamma_kG/\gamma_{k+1}G \otimes \Bbb Q$$ to Malcev algebra which becomes isomorphism after taking associated graded quotients, as proven by Quillen in paper On the associated graded ring of a group ring, 1968.

We have a theorem due to Sullivan.

$$\square$$ Space $$X$$ is $$1$$-formal if and only if degree completion of holonomy Lie algebra $$hol(X)$$ is filtered isomorphic to $$\mathfrak m (\pi_1(X))$$. $$\square$$

Good reference is two books by Felix, Halperin, Thomas, called (unsurprisingly) Rational homotopy theory I/II. 1-formality is very clearly explained in the beginning of second tome.

Now, to your question. If a space is formal, it is indeed $$q$$-formal for all $$q$$, in particular, 1, so your (3) implies both (2) and (1).

YCor mentioned in comments that (1) is strictly weaker as witnessed by integral Heisenberg group. More generally, every two-step nilpotent Lie algebra is isomorphic to its associated lower central quotient, but only virtually abelian nilpotent groups are 1-formal. (If some commutator $$[a, b]$$ is nontrivial and there are no relations in weight 3, then $$[a, [a, b]]$$ will be nontrivial).

Let's prove that (2) implies (3).

Suppose $$\mathfrak m(G)$$ is a completion of some quadratic algebra $$E = FreeLie(V)/(R), R \subset \wedge^2 V$$. Now, going back to Quillen, we have a homomorphism $$L(G) \to \mathfrak m(G)$$ which becomes isomorphism on associated graded (in particular, it's injective). As Malcev algebra was graded from the beginning, $$L(G)$$ inherits this grading, and the only thing it can be is $$E$$. From Sullivan's Infinitesimal computations in topology we know that kernel of Lie bracket induced by group commutator on $$G_{ab} \wedge G_{ab}$$ rationally isomorphic to image of comultiplication $$\mu^*$$, so $$E \cong hol(G)$$ q.e.d.