# Period map for $\partial\bar\partial$-manifolds

When we talk about the theory of variation of Hodge structures, we always assume that the central fiber is a Kähler manifold $$X$$, then consider a family of deformations $$\pi:\mathcal X\to B$$ and the period map $$\mathcal P:B\to Grass(b^{p,k},H^k(X,\mathbb C))$$, $$b^{p,k}=dim F^pH^k(X,\mathbb C)$$.

What if we replace the Kähler manifold by a $$\partial\bar\partial$$-manifold and consider a holomorphic family of $$\partial\bar\partial$$-manifolds $$\pi:\mathcal X\to B$$?

Recall a $$\partial\bar\partial$$-manifold is a compact complex manifold which satisfies for any $$\partial$$, $$\bar\partial$$-closed $$d$$ exact $$(p,q)$$ form $$\alpha$$, there this a $$(p-1,q-1)$$ form $$\beta$$ such that $$\alpha=\partial\bar\partial \beta$$.

Is the theory of variation of Hodge structures remains the same? And is there also a period map $$\mathcal P:B\to Grass(b^{p,k},H^k(X,\mathbb C))$$? If so, is the period map holomorphic and the Griffiths transversality still holds?

In my opinion, first we should define what $$F^pH^k(X,\mathbb C)$$ means for a non-Kähler manifold, since for a Kähler manifold, we have the Kähler identity $$\Delta=2\Delta_{\bar\partial}=2\Delta_{\partial}$$ from which we deduce the Hodge decomposition: $$H^k(X,\mathbb C)=\oplus_{p+q=k}H^{p,q}(X)$$, so we can define $$F^pH^k(X,\mathbb C)=H^{p,k-p}(X)\oplus H^{p+1,k-p-1}(X)\oplus...\oplus H^{k,0}(X)$$ which is obviously a subspace of $$H^k(X,\mathbb C)$$, but for a non-Kähler manifold, there is no Kähler identities. But from AT13, we know that the Bott-Chern cohomology $$H^{p,q}_{BC}(X,\mathbb C):=\frac{ker\partial\cap ker\bar\partial}{im\partial\bar\partial}$$ is isomorphic to Dolbeault cohomology $$H_{\bar\partial}^{p,q}(X)$$ and $$H_{\partial}^{p,q}(X)$$, so if we define $$F^pH^k(X,\mathbb C)=H^{p,k-p}_{BC}(X,\mathbb C)\oplus H^{p+1,k-p-1}_{BC}(X,\mathbb C)\oplus...\oplus H^{k,0}_{BC}(X,\mathbb C)$$, is it reasonable to treat is as a subspace of $$H^k(X,\mathbb C)$$ and define a period map as in the Kähler case?

Let me start with a disclaimer that I think the following facts are true, but I'm doing this over coffee and I haven't checked the details carefully. First, I'll redefine $$F^pH^k(X,\mathbb{C})$$ to be the space of de Rham classes represented by the sum of $$(p', p'-k)$$ forms with $$p'\ge p$$, or equivalently as $$F^pH^k(X,\mathbb{C})= im(H^k(\Omega_X^{\ge p})\xrightarrow{\iota} H^k(X,\Omega_X^\bullet))$$ Then the $$\partial\bar\partial$$-lemma is sufficient to guarantee the filtration is strict in the sense that the maps above are injective (or equivalently that the Hodge to de Rham spectral sequence degenerates); edit see remark 5.21 of Deligne, Griffiths, Morgan, Sullivan, Real homotopy theory of Kähler manifolds. Then, if I understand the notation of the paper you linked, this should probably give decomposition in terms of BC cohomology as you wrote. The other thing I want to remark is that if $$\mathcal{X}\to B$$ is a smooth proper family such that the fibres satisfy $$\partial\bar\partial$$-lemma, then usual arguments should imply that $$F^p R^kf_*\Omega_{\mathcal{X}/B}^\bullet= im(R^kf_*\Omega_{\mathcal{X}/B}^{\ge p}\xrightarrow{\iota}R^kf_*\Omega_{\mathcal{X}/B}^\bullet)$$ satisfies Griffiths transversality etc. So in this sense, things work. However, you would be missing the polarization, which need for most of the deeper results about the Griffiths period map.
• Do you mean you define $F^pH^k(X,\mathbb C)=ker(d:F^pA^k\to F^{p+1}A^k)/im(d:F^{p-1}A^k\to F^pA^k)$?
• Almost, but there is no shift because $d(F^p)\subset F^p$ Jul 24 at 14:50
• Sorry that I made a mistake, I mean $F^pH^k(X,\mathbb C)=ker(d:F^pA^k\to F^pA^{k+1})/im(d:F^pA^{k-1}\to F^pA^k)$. If we define the filtration like this, I don't think it obvious that we will have of decompostion of this filtration in BC cohomology, since for the Kahler case, it takes p158-159 for Voisin in her book<Hodge theory and...> to prove the corresponding decomposition.
• Technically, it gives an isomorphism $F^pH^k(X)= H^{p,k-p}\oplus\ldots$ with Dolbeault cohomology. Now use the fact, you stated, that BC and Dolbeault cohomologies are isomorphic. Jul 25 at 14:36