**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.