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 it as a subspace of $H^k(X,\mathbb C)$ and define a period map as in the Kähler case?

Has anyone thought it before? or any reference about this issue?


1 Answer 1


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.

  • $\begingroup$ Do you mean you define $F^pH^k(X,\mathbb C)=ker(d:F^pA^k\to F^{p+1}A^k)/im(d:F^{p-1}A^k\to F^pA^k)$? $\endgroup$
    – Tom
    Commented Jul 24, 2021 at 14:24
  • $\begingroup$ Almost, but there is no shift because $d(F^p)\subset F^p$ $\endgroup$ Commented Jul 24, 2021 at 14:50
  • $\begingroup$ 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. $\endgroup$
    – Tom
    Commented Jul 24, 2021 at 15:14
  • 1
    $\begingroup$ Voisin's argument assumes the Kahler condition, so it won't work here. But see the ref. in the edited version. $\endgroup$ Commented Jul 24, 2021 at 18:42
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    $\begingroup$ 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. $\endgroup$ Commented Jul 25, 2021 at 14:36

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