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.

  • 1
    $\begingroup$ Nice question. Perhaps the chapter on Sullivan algebras in the book of Felix, Halperin and Thomas does what you want? There are some simple examples there, but I don't have my copy here to check. $\endgroup$
    – Mark Grant
    Jan 2, 2011 at 19:13
  • $\begingroup$ Mark -- thanks. I don't have a copy of that book at hand either, but I don't recall having seen an example there. $\endgroup$
    – algori
    Jan 3, 2011 at 0:34

3 Answers 3


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 $\langle \nu_{123}, \nu_{234}, \nu_{123}\rangle$...

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    $\begingroup$ That is a very cool example! $\endgroup$ Jan 3, 2011 at 17:35

Edit: Let me give it one more try -- I think I've fixed the example.

The following is the simplest simply connected example I know (note that André's example has a large fundamental group).

Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,y,z, a_1,a_2,b_1,b_2, c_1,c_2,\ldots\rangle,d)$ with $\deg u = \deg v = \deg w = 3, \deg x = \deg y = \deg z =8, \deg a_i = \deg b_i =\deg c_i= 10$ and the following differentials:

$ du=dv=dw=0, dx=uvw, dy=dz=0, da_1=xu+yv, da_2=xu+zw, db_1=xv+yw, $

$ db_2=xv+zu, dc_1=xw+yu, dc_2=xw+zv $

Keep adding generators to kill off all cohomology in degrees above 11. I believe this algebra finally has all Massey products equal to zero. For degree reasons the only possible nontrivial Massey products can be for two classes of degree 3 and one of degree 6, e.g. something of the form $\langle [u], [vw], [v]\rangle$. As far as I can tell all of them vanish.

On the other hand this algebra is not formal. Indeed, suppose we have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(\Lambda V, d),0)$. Let $I=\ker \phi$. Since $\phi$ is a quasi-isomorphism, every closed element of $I$ is exact. Now, since $[y]$ and $[z]$ are independent cohomology classes of degree 8 we must have that $I^{\le 8}=\langle x+k_1y+k_2z\rangle$ for some $k_1,k_2\in\mathbb Q$. Then the class $[u(x + k_1y+k_2z)] = [ux] + k_1[u][y]+k_2[uz]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal.

Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.


Many examples of closed highly-connected manifolds with all Massey products vanishing, yet which are not formal, are given in "The rational homotopy type of $(n-1)$-connected manifolds of dimension up to $5n-3$" by Crowley and Nordström. See e.g. Example 1.11: https://arxiv.org/abs/1505.04184.


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