# When a symplectic manifold is formal?

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?

• What is the Lie bracket here? – user36931 Apr 3 '14 at 14:13
• @user: perhaps the OP is using the symplectic form to identify forms and polyvector fields, then using the Schouten bracket? – Qiaochu Yuan Apr 4 '14 at 17:56
• I think this is the symplectic bracket, not the transferred Schouten bracket. – Gabriel C. Drummond-Cole 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 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.
• 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). Dotsenko-Shadrin-Vallette 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. – Gabriel C. Drummond-Cole Apr 5 '14 at 10:41
• 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. – Gabriel C. Drummond-Cole Apr 5 '14 at 10:44