Let $(M,\omega)$ be a $2n$dimensional symplectic manifold, then we have the symplectic Hodge operator $$*:\Omega^{k}(M)\rightarrow\Omega^{2nk}(M)$$ 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 GerstenhaberBatalinVilkovisky 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 quasiisomorphic to the cohomology $H^{*}_{dR}(M)$ which is regarded as a DGLA with the trivial differential and the trivial Lie bracket?

$\begingroup$ What is the Lie bracket here? $\endgroup$ – user36931 Apr 3 '14 at 14:13

$\begingroup$ @user: perhaps the OP is using the symplectic form to identify forms and polyvector fields, then using the Schouten bracket? $\endgroup$ – Qiaochu Yuan Apr 4 '14 at 17:56

$\begingroup$ I think this is the symplectic bracket, not the transferred Schouten bracket. $\endgroup$ – Gabriel C. DrummondCole Apr 5 '14 at 10:38
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 dGBValgebra. 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.

$\begingroup$ In more generality, as long as $d^*$ is homotopically trivial, which is a significantly weaker condition that doesn't require Hard Lefschetz, the Lie algebra is formal (and zero on homology). DotsenkoShadrinVallette have some results on geometric conditions that imply this algebraic criterion in the symplectic case (and generalizations) although I'm not sure whether that work is published yet. $\endgroup$ – Gabriel C. DrummondCole Apr 5 '14 at 10:41

$\begingroup$ Also, $d^*$ being homotopically trivial is sufficient but at least algebraically it is not necessary; it's easy to come up with a dgBV algebra with homotopically nontrivial $d^*$ but formal induced Lie algebra equivalent to $0$ on homology (for example, if the product is zero). I don't know whether such examples can be realized geometrically. $\endgroup$ – Gabriel C. DrummondCole Apr 5 '14 at 10:44