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?

  • 2
    $\begingroup$ Clearly 2 implies 1, and the converse doesn't hold (for instance the 3-dimensional Heisenberg Lie algebra satisfies 1 and not 2). $\endgroup$
    – YCor
    Jan 16, 2020 at 22:00
  • 3
    $\begingroup$ 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 $\endgroup$
    – Adrien
    Jan 17, 2020 at 9:21

1 Answer 1


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.


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