# Are there geometrically formal manifolds, which are not rationally elliptic?

Formality of a space is meant in the sense of Sullivan, i.e. a space $X$ is called formal, if its commutative differential graded algebra of piecewise linear differential forms $(A_{PL}(X),d)$ is weakly equivalent (There is zig-zag of quasi-isomoprhisms.) to the rational cohomology algebra $(H^*(X,\mathbb{Q}),0)$ of $X$. Since I'm only interested in manifolds, this is equivalent to the case of weak equivalence between the de Rham algebra of differential forms $(A_{DR}(X),d)$ and the real cohomology algebra of $X$.

A compact oriented manifold $X$ is called geometrically formal, if it admits a formal metric, i.e. a Riemannian metric $g$, such that the wedge product of harmonic forms is harmonic again. The typical class of examples are compact symmetric spaces.

By using Hodge decomposition, it's easy to see, that geometric formality implies formality.

The converse is not true, geometric formality is much stronger than formality. A formal metric has topological consequences, which go beyond formality. One can for example show that the Betti numbers of a geometric formal manifold $X$ are bounded above by the one of a torus in the same dimension, e.g. $$b_i(X)\le b_i(T^{dim(X)})={dim(X) \choose i}.$$

This bound on the Betti numbers is also known for rationally elliptic spaces $X$, that are spaces whose total dimensions of their rational cohomology $H^*(X,\mathbb{Q})$ and their rational homotopy $\pi_*(X)\otimes\mathbb{Q}$ are finite. For compact manifolds, this is clearly equivalent to having finite dimensional rational homotopy.

There are examples of rationally elliptic compact simply-connected manifolds, which are not geometrically formal. (Some of them generalized symmetric spaces, see On formality of generalised symmetric spaces by D. Kotschick and S. Terzic)

Are there examples of compact oriented manifolds, which are geometrically formal, but not rationally elliptic? Are there even examples of simply connected ones?

Edit: As a sidenote, beside symmetric spaces, one can show, that manifolds with nonnegative curvature operatore are geometric formal. But since those have nonnegative sectional curvature, they won't serve as counterexamples easily (at least in the simply connected case), because of the Bott conjecture, which states that all simply connected nonnegative sectional curved manifolds are rationally elliptic.

• Note that examples of these kind cannot be homogeneous spaces or even biquotients, as they are rationally elliptic. Commented Mar 11, 2014 at 20:43
• Is geometric formality preserved under rational homotopy equivalence? Commented Mar 12, 2014 at 3:13
• @IgorBelegradek In case you're still interested: It is not; by Corollary 11 in Kotschick's original paper arxiv.org/abs/math/0004009, a closed oriented four-manifold with $b_1=1$ and $b_2=0$ is geometrically formal iff it fibers over a circle. Now, $S^3\times S^1$ is geometrically formal, but if you connect sum that with a non-trivial rational homology four-sphere, then you get a manifold with the same rational homotopy type, but which does not fiber over a circle. (This last part, about not fibering over a circle, was shown to me by Corey Bregman; I can provide details if you're interested.) Commented Jun 9, 2019 at 18:54

Consider complex conjugation $\mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2$. Perturb this map by taking an affine chart $\mathbb{C}^2$ and setting $f(z) = \rho(|z|)\cdot \bar{z}$, where $\rho$ is a smooth map to $\mathbb{R}_{\geq 0}$ that is 0 for $|z| < 1$, and 1 for $|z| > 2$, say. The map $f\colon \mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2$ is orientation-preserving and its effect on $H^2(\mathbb{C}\mathbb{P}^2)$ is that of negative the identity. Now, take two copies of $\mathbb{C}\mathbb{P}^2$, cut out small neighborhoods of the balls in the corresponding affine charts where $\rho$ makes the map $f$ constant, and glue the resulting manifolds together to obtain $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ along with an orientation-preserving diffeomorphism $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ whose effect on second cohomology is given by $(\begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix})$. Repeat this procedure on $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ and $\mathbb{C}\mathbb{P}^2$ to obtain an orientation-preserving diffeomorphism $g \colon (\mathbb{C}\mathbb{P}^2)^{\#3} \to (\mathbb{C}\mathbb{P}^2)^{\#3}$ whose induced map on second cohomology is $$\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}.$$ Now consider the mapping torus $M$ of $g$. As can be seen from the long exact sequence in homotopy for the fibration $(\mathbb{C}\mathbb{P}^2)^{\#3} \to M \to S^1$, the rational homotopy $\pi_*(M) \otimes \mathbb{Q}$ has infinite rank. From the long exact sequence in relative cohomology for the inclusion $(\mathbb{C}\mathbb{P}^2)^{\#3} \hookrightarrow M$ we see that $H^2(M, \mathbb{Q}) = 0$. Therefore $M$ has Betti numbers $b_1 = 1, b_2 = 0, b_3 = 0, b_4 = 1$, and so by Theorem 8 in Kotschick's On products of harmonic forms'' (https://arxiv.org/abs/math/0004009) it admits a formal metric.
Note that the manifold $M$ has the rational cohomology ring of $S^1 \times S^4$ and so, since geometric formality implies formality, the minimal model of $M$ is given by $$\Lambda(x_1, x_4, y_7; \ \ dx_1 = dx_4 = 0, \ dy_7 = x_4^2),$$ where subscripts denote degrees. It would then seem from this model that $M$ is in fact rationally elliptic. However, $M$ is not a nilpotent space (consider the action of $\pi_1$ on $\pi_2$), and so the ranks of the rational homotopy groups cannot be read off immediately from the amount of generators in the model. Also note that "rationally elliptic" is a concept that, at least in Félix, Halperin, Thomas, is defined only for simply connected spaces. So the above example doesn't truly answer your question.