7
$\begingroup$

It is known that for a $\partial\bar\partial$-manifold $X$ (a compact complex manifold satisfies the $\partial\bar\partial$-lemma), the Bott-Chern cohomology $H_{BC}^{\bullet,\bullet}:=\frac{\ker \partial\cap\ker\bar\partial}{\text{im }\partial\bar\partial}$ is isomorphic to the Dolbeault cohomology $H_{\bar\partial}^{\bullet,\bullet}(X)=\frac{\ker\bar\partial}{\text{im }\bar\partial}$, and the Frölicher spectral sequence of $X$ degenerates at $E_1$, then $$ H^k(X,\mathbb C)=\bigoplus_{p+q=k}H^{p,q}_{\bar\partial}(X)=\bigoplus_{p+q=k}H^{p,q}_{BC}(X), $$ so $H^{p,q}_{\bar\partial}(X)$ can be treated as a subspace of $H^k(X,\mathbb C)$ through the isomorphism between Bott-Chern and Dolbeault cohomology.

If we only assume that the Frölicher spectral sequence of $X$ degenerates at $E_1$, is there a natural way to treat $H^{p,q}_{\bar\partial}(X)$ as a subspace of $H^k(X,\mathbb C)$? or is there a natural map $f:H^{p,q}_{\bar\partial}(X)\to H^k(X,\mathbb C)$ which is injective?

Improve this question
$\endgroup$
13
  • $\begingroup$ Every surface has Fr\"olicher spectral sequence degenerating at $E_1$, so there is probably no chance to have any natural map $\endgroup$
    – Grisha Papayanov
    Commented Jun 8, 2023 at 16:08
  • $\begingroup$ Noncanonically, though, one can perform the following: choose a splitting of the cohomology of the $\bar\partial$ operator. That is, pick a representative for any $\bar\partial$-cohomology class, choose the complement of the space of the representatives inside all forms and choose an operator $h$ of degree $-1$ such that $\mathrm{Id} - \partial h - h\partial$ is the corresponding projection onto $\bar\partial$-cohomology. Then, if $\alpha$ is a $\bar\partial$-closed $(p,q)$-form, the form $\sum_{n \ge 0}(h\partial)^n\alpha$ will be $d$-closed and non-exact if FSS degenerates at $E_1$. $\endgroup$
    – Grisha Papayanov
    Commented Jun 8, 2023 at 16:17
  • $\begingroup$ This follows, for example, from the Khovanov's classification of bicomplexes into zig-zags: arxiv.org/abs/1911.02503. The Fr\"olicher spectral sequence degenerates at $E_1$ if and only if there are no zig-zags of odd length in the corresponding bicomplex. $\endgroup$
    – Grisha Papayanov
    Commented Jun 8, 2023 at 16:23
  • $\begingroup$ sorry, it should read "if $\alpha$ is a $\bar\partial$-closed and not $\bar\partial$-exact $(p,q)$-form..." $\endgroup$
    – Grisha Papayanov
    Commented Jun 8, 2023 at 16:26
  • $\begingroup$ @GrishaPapayanov, your construction of $\mathrm{Id} - \partial h - h\partial$ seems very unnatural, I don't why we should care about this kind of operator. $\endgroup$
    – Tom
    Commented Jun 9, 2023 at 3:33

1 Answer 1

Reset to default
1
+100
$\begingroup$

I'm not sure that the following answers the question, but at least it shows that it is not always possible to construct a map $H_{\bar\partial}\to H_{dR}$ by assigning a $d$-closed representative to each class in $H_{\bar\partial}$.

Suppose $\alpha$ is a $\bar\partial$-closed form, and we want to find $\beta$ such that $d(\alpha-\bar\partial\beta)=0$. Each summand inside the parentheses is $\bar\partial$-closed, so the equality is equivalent to $\partial\alpha=\partial\bar\partial\beta$. We learn that each class in $H_{\bar\partial}$ admits a $d$-closed representative if and only if $\partial\ker\bar\partial\subset\operatorname{im}\partial\bar\partial$ — in a more fancy language, the map $H_{\bar\partial}\to H_{BC}$ induced by $\partial$ is zero.

As explained in Ornea, Verbitsky, Vuletescu, Classification of non-Kähler surfaces and locally conformally Kähler geometry, § 2.2, for any non-Kähler compact complex surface, the image of $H^{0,1}_{\bar\partial}$ by this map is 1-dimensional. So there is a class in $H^{0,1}_{\bar\partial}$ with no $d$-closed representative, notwithstanding that the Frölicher spectral sequence of any compact complex surface degenerates at the first page.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .