Let $X$ be a polyhedron. For each $n$-dimensional face $f$ of $X$ fix a homeomorphism $\sigma_f:\triangle^n\to f$ where $\triangle^n$ is the standard $n-$simplex so that whenever $f$ is a face of $f'$ the map $\sigma_{f'}^{-1}\sigma_f$ is an affine map $\triangle^n\to\triangle^{n'}$ where $n'=\dim \triangle'$. These maps enable one to speak of forms on the simplices of $X$ with rational polynomial coefficients. A degree $n$ Sullivan piecewise polynomial form on $X$ is a choice of degree $n$ rational polynomial forms, one for each face of $X$, that agree on the intersections of the faces. Such forms form a commutative differential graded algebra (cdga) denoted $A_{PL}(X)$.
$X$ is called formal if the cdga's $A_{PL}(X)$ and $H^\ast(X,\mathbf{Q})$ (with zero differential) are quasi-isomorphic i.e. can be connected by a chain of cdga quasi-isomorphisms. One topological consequence of formality is that if $X$ is formal, all higher Massey products on $H^\ast(X,\mathbf{Q})$ vanish (see e.g. Griffiths, Morgan, Rational homotopy theory and differential forms). I vaguely remember having heard that the converse is false: there are non-formal spaces with vanishing Massey products. I would like to ask if anyone knows a (preferably not too expensive) example of such a space.
Remark: this question can of course be stated in terms of cdga's without referring to spaces: by Sullivan's realization theorem (Infinitesimal computations in topology, \S 8) to give an example it would suffice to construct a non-formal cdga $A$ with $A^0=\mathbf{Q},A^1=0$ all of whose Massey products vanish. However, if there is a geometric example, I'd be interested to know.
Best Answer
The real locus of the moduli space of stable curves of genus zero with $6$ marked points is a non-formal space with vanishing Massey products. This is a compact non-orientable 3-manifold that is hyperbolic on the complement of $10 = \frac{1}{2} { 6 \choose 3 }$ embedded tori.
See the remark on page 4 of my paper with Etingof, Kamnitzer, and Rains.
The rational cohomology algebra of that manifold is generated by symbols $\nu_{ijk}$ for $1\le i\lt j\lt k\le 5$, and has defining relations given by $$ \nu_{ijk}\nu_{ijl}=0. $$ See Proposition 2.3 of the above paper (the second relation in loc. cit. doesn't occur because $6$ is too small).
The Massey products vanish because
their indeterminacy is too big: there is no room for them to be non-zero..... wait, I need to check this.I forget if there is a good argument about why the Massey products should vanish. But I redid the computation $\langle \nu_{123}, \nu_{234}, \nu_{345}\rangle$ and found zero. The way I did the computation was by representing the cohomology classes by explicit submanifolds, and doing intersections.
More precisely:
• $\nu_{123}$ is represented by the submanifold $M_{123}$ where the marked points "0" and "1" are separated from the marked points "2" and "3" by a normal crossing.
• $\nu_{234}$ is represented by the submanifold $M_{234}$ where the marked points "0" and "2" are separated from the marked points "3" and "4" by a normal crossing.
• $\nu_{345}$ is represented by the submanifold $M_{345}$ where the marked points "0" and "3" are separated from the marked points "4" and "5" by a normal crossing.
At this point, one needs to check that the manifolds $M_{123}$ and $M_{234}$ are transverse, and that the manifolds $M_{234}$ and $M_{345}$ are transverse (a necessary condition for this kind of computation to work out).
• The intersection $M_{123}\cap M_{234}$ is bounded by $\frac {1} {2}$ times the manifold where the points "0", "1", "2" are separated from the points "3", "4". Call that manifold $A$.
• The intersection $M_{234}\cap M_{345}$ is bounded by $\frac {1} {2}$ times the manifold where the points "0", "2" are separated from the points "3", "4", "5". Call that manifold $B$.
Finally, we need to compute the cohomology class of $X := (M_{123}\cap B) \cup (A \cap M_{234})$. That manifold represents the Massey product $\langle \nu_{123}, \nu_{234}, \nu_{345}\rangle$. Unfortunately, $X$ is not empty. But you can intersect it with all the generators of $H_2$ and see that you always get zero. The generators of H_2 are the tori I talked about in the beginning of my answer. The manifold $X$ is actually disjoint from all but one of these tor, and intersects that last torus non-transversely. Even worse: it is contained in that torus. One then needs to deform it and see that it can be made disjoint from that last torus.
I guess I should also check the Massey product $\langle \nu_{123}, \nu_{234}, \nu_{123}\rangle$...