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Let $X$ be a complex manifold, there is a natural map $f:H^2(X,\mathbb C)\to H^2(X,\mathcal O)$ induced by the inclusion map $\mathbb C\hookrightarrow \mathcal O$ which coincides with the natural projection map $\pi^{0,2}:H^2_{dR}(X,\mathbb C)\to H^2(X,\mathcal O),[\omega]\mapsto [\omega^{0,2}]$, where $\pi^{0,2}:A^2(X)\to A^{0,2}(X)$ take a $2$-form to its $(0,2)$ component (see Huybrechts' book complex geometry: an introduction p.132 Lemma 3.3.1).

When $X$ is a compact Kähler manifold, it is easy to conclude that the natural map $f:H^2(X,\mathbb C)\to H^2(X,\mathcal O)$ is surjective. Since for any $\Delta_{\bar\partial}$-harmonic representative $\omega^{p,q}$ of $H^{p,q}_{\bar\partial}(X)$, we can add $\omega^{p,q}$ together and get $\omega=\sum\limits_{p+q=k}\omega^{p,q}$, which is $\Delta$-harmonic by Kähler identity: $\Delta=2\Delta_{\bar\partial}$.

But for a non-Kähler manifold, for example, the $\partial\bar\partial$-manifold: a compact complex manifold with any $\partial$-,$\bar\partial$-closed, $d$-exact $(p,q)$ form being $\partial\bar\partial$-exact, is the map $f$ surjective?

I guess that this map is also surjective. Since its Frolicher spectral sequence degenerates at $E_1$ (see remark 5.21 of Deligne, Griffiths, Morgan, Sullivan, Real homotopy theory of Kähler manifolds), so we have $H^2(X)\cong H^{2,0}_{\bar\partial}(X)\oplus H^{1,1}_{\bar\partial}(X)\oplus H^{0,2}_{\bar\partial}(X)$, and there is an isomorphism between Bott-Chern cohomology and Dolbeault cohomology $H_{BC}^{p,q}(X)\cong H^{p,q}_{\bar\partial}(X)$ for any $\partial\bar\partial$-manifold. But for any $\bar\partial$-close $(0,2)$ form $\alpha$, how can we conclude there must be a $d$-closed form $\omega$ with its $(0,2)$ component $\omega^{0,2}$ equals to $\alpha$?

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    $\begingroup$ You more or less answered your own question. If the $\partial\overline{\partial}$-lemma holds then the Frölicher (or Hodge to de Rham) spectral sequence degenerates. This implies that the so called edge map $H^p(X,\mathbb{C})\to H^p(X,\mathcal{O})$ is surjective. This can be made more explicit. $\endgroup$ Commented Oct 3, 2022 at 13:14
  • $\begingroup$ @Donu Arapura, According to your comment, I guess we only need the $(0,2)$ part of the condition of Frölicher spectral sequence degenerates at $E_1$, that is $H_{\bar\partial}^{0,2}(X)=\frac{H^2(X)}{F^1H^2(X)}$, where $F^1H^2(X)$ means $d$-closed $F^1A^2$ modulo $d$-exact ones in $F^1A^2$, so there is $\frac{Z_{\bar\partial}^{0,2}}{B_{\bar\partial}^{0,2}}=\frac{Z^2}{B^2}/\frac{F^1Z^2}{F^1B^2}$, by taking projection $\pi^{0,2}$, there is for any $\bar\partial$-closed $(0,2)$ form $\omega^{0,2}$, there is a closed $2$ form $\omega$ whose $(0,2)$ component is $\omega^{0,2}$? $\endgroup$
    – Tom
    Commented Oct 4, 2022 at 10:12

2 Answers 2

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For compact complex surfaces (i.e., when $\dim X =2$) the map is always surjective. See Theorem 2.10, p.141 in

Barth, Wolf P.; Hulek, Klaus; Peters, Chris A. M.; Van de Ven, Antonius, Compact complex surfaces, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge 4. Berlin: Springer (ISBN 3-540-00832-2/hbk). xii, 436 p. (2004). ZBL1036.14016.

In general, the $\partial \bar{\partial}$-lemma (Lemma 13.6 p. 44 of the reference above) says that, if $X$ is a compact complex manifold such that

  1. the Frölicher spectral sequence degenerates at $E_1$ and
  2. there is a formal Hodge decomposition,

then $H^{p, \, q}(X)$ coincides with the subspace of $H^{p+q}(X)$ representable by $d$-closed forms of type $(p, \, q)$.

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  • $\begingroup$ I still have 2 questions: (i) Is your condition (2) necessary to ensure the map $f$ being surjective? which seems contradict to Prof Arapura's comment. (ii) Your condition (1): $E_1^{p,q}=E_{\infty}^{p,q}$, which means $H^{p,q}_{\bar\partial}(X)=\frac{F^pH^{p+q} (X)}{F^{p+1}H^{p+q}(X)}$, where $F^pH^{p+q}:=\frac{F^pA^{p+q}\cap\text{ker }d}{F^pA^{p+q}\cap\text{im }d}$, but both of these two groups are not the subspace of $H^{p+q}(X)$ represented by $d$-closed forms, which may be seemed as $\frac{A^{p,q}\cap\text{ker }d}{A^{p,q}\cap\text{im }d}$? Can you elaborate it a bit? $\endgroup$
    – Tom
    Commented Oct 4, 2022 at 9:54
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According to @Donu Arapura's comment, I give an answer of my understanding, whether it is correct or not, please comment below.

By the definition of Frölicher spectral sequence degenerating at $E_1$, we have particularly $E_1^{0,2}=E_{\infty}^{0,2}$, which is equivalent to $H^{0,2}_{\bar\partial}(X)=\frac{H^2(X,\mathbb C)}{F^1H^2(X,\mathbb C)}$, where $F^1H^2(X,\mathbb C):=\frac{F^1A^2(X)\cap\ker d}{F^1A^2(X)\cap\text{im }d}$. Then the map $H^2(X,\mathbb C)\to H^2(X,\mathcal O)$ becomes $H^2(X,\mathbb C)\to \frac{H^2(X,\mathbb C)}{F^1H^2(X,\mathbb C)}$, which seems obviously surjective.

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