TITLE: When a symplectic manifold is formal?
QUESTION [7 upvotes]: Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold, then we have the symplectic Hodge operator 
$$*:\Omega^{k}(M)\rightarrow\Omega^{2n-k}(M)$$
Furthermore, we can define a differential $d^{*}=(-1)^{k+1}*d*$ which acts on $\Omega^{k}(M)$. A canonical result of symplectic geometry say that $(\Omega^{*}(M),d,d^{*})$ forms a differentiable Gerstenhaber-Batalin-Vilkovisky algebra (dGBV). The structure of dGBV induces a differential graded Lie algebra (DGLA) on $\Omega^{*}(M)$. So my question is when the de Rham complex of a symplectic manifold is a formal DGLA, i.e. $\Omega^{*}(M)$ is quasi-isomorphic to the cohomology $H^{*}_{dR}(M)$ which is regarded as a DGLA with the trivial differential and the trivial Lie bracket?

REPLY [2 votes]: There is a paper of Merkulov (http://arxiv.org/abs/math/9805072), where he proves that the Hard Lefschetz condition is equivalent to the "$dd^*$-lemma", condition that $\mathrm{Im}\ d \cap \mathrm{Ker}\ d^* = \mathrm{Im}\ d \cap \mathrm{Ker}\ d^* = \mathrm{Im}\ dd^*$. 
Though it is not clear what is meant by formality in the paper, but i heard from him that it is formality of dGBV-algebra. It shouldn't be hard to deduce dGBV- (and dGLA-) formality from $dd^*$-lemma, though I haven't checked it with pen and paper, so sorry if I am mistaken.