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 $\alpha$?

  • 1
    $\begingroup$ 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


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 $(p, \, q)$.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.