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In Angella and Tomassini's paper p.75, there is an exact sequence:

$\cdots\to B^{\bullet,\bullet}\to H_{\bar\partial}^{\bullet,\bullet}\to H_A^{\bullet,\bullet}\to \cdots$

where $B^{\bullet,\bullet}:=\frac{\ker\bar\partial\cap\text{im }\partial}{\text{im }\partial\bar\partial}$, and recall that Dolbeault cohomology $H_{\bar\partial}^{\bullet,\bullet}:=\frac{\ker\bar\partial}{\text{im }\bar\partial}$, and Aeppli cohomology $H_A^{\bullet,\bullet}:=\frac{\ker\partial\bar\partial}{\text{im }\partial+\text{im }\bar\partial}$, obviously, there are natural maps of $f:B^{\bullet,\bullet}\to H_{\bar\partial}^{\bullet,\bullet}$ and $g:H_{\bar\partial}^{\bullet,\bullet}\to H_A^{\bullet,\bullet}$, but I wonder how do we get $\text{im } f=\ker g$?

My thinking process is like that: In order to get the expression of $B^{\bullet,\bullet}$, we should compute the kernel of the map $g$, for a $\bar\partial$-closed form $\alpha$, let $g(\alpha)=0\in H_A^{\bullet,\bullet}$, then we get $\alpha=\partial\beta+\bar\partial\gamma$, since $\alpha$ is $\bar\partial$-closed, we get $\bar\partial\partial\beta=0$, thus $\alpha\in\ker\bar\partial\cap\text{im }\partial+\text{im }\bar\partial$, then I get stuck, can anyone help me to get $B^{\bullet,\bullet}=\frac{\ker\bar\partial\cap\text{im }\partial}{\text{im }\partial\bar\partial}$?

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Let $\alpha$ be a $\bar{\partial}$-closed form. Denote its Dolbeault cohomology class by $[\alpha]_{\bar{\partial}}$ and its Aeppli cohomology class by $[\alpha]_A$; note that the map $g$ is given by $g([\alpha]_{\bar{\partial}}) = [\alpha]_A$. Likewise, if $\alpha' \in \ker\bar{\partial}\cap\operatorname{im}\partial$, denote the corresponding element in $B^{\bullet,\bullet}$ by $[\alpha']_B$; note that the map $f$ is given by $f([\alpha']_B) = [\alpha']_{\bar{\partial}}$.

As you noted, if $[\alpha]_{\bar{\partial}} \in \ker g$, then $\alpha = \partial\beta + \bar{\partial}\gamma$, so $[\alpha]_{\bar{\partial}} = [\partial\beta + \bar{\partial}\gamma]_{\bar{\partial}} = [\partial\beta]_{\bar{\partial}}$. Moreover, since $\bar{\partial}\alpha = 0$, we see that $\bar{\partial}\partial\beta = 0$ and hence $\partial\beta \in \ker\bar{\partial}\cap\operatorname{im}\partial$. Therefore, we can form the element $[\partial\beta]_B$ which satisfies $f([\partial\beta]_B) = [\partial\beta]_{\bar{\partial}} = [\alpha]_{\bar{\partial}}$, so $[\alpha]_{\bar{\partial}} \in \operatorname{im}f$ and hence $\ker g \subseteq \operatorname{im}f$.

Suppose now that $[\alpha]_{\bar{\partial}} \in \operatorname{im}f$, then there is $\alpha' \in \ker\bar{\partial}\cap\operatorname{im}\partial$ with $[\alpha']_{\bar{\partial}} = [\alpha]_{\bar{\partial}}$. As $\alpha' \in \operatorname{im}\partial + \operatorname{im}\bar{\partial}$, we see that $g([\alpha]_{\bar{\partial}}) = g([\alpha']_{\bar{\partial}}) = [\alpha']_A = 0$, so $[\alpha]_{\bar{\partial}} \in \ker g$ and hence $\operatorname{im}f \subseteq \ker g$.

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  • $\begingroup$ The explanation is precise! But why the denominator of $B^{\bullet,\bullet}$ should be $\text{im }\partial\bar\partial$? $\endgroup$
    – Tom
    Sep 15 at 3:07
  • $\begingroup$ Your question is about exactness at $H^{\bullet,\bullet}_{\bar{\partial}}$. Replacing $f : B^{\bullet,\bullet} \to H^{\bullet,\bullet}_{\bar{\partial}}$ with any map which has the same image will preserve exactness at $H^{\bullet,\bullet}_{\bar{\partial}}$. That is, exactness at $H^{\bullet,\bullet}_{\bar{\partial}}$ doesn't force the previous term to be $B^{\bullet,\bullet}$. I assume that the specific form of $B^{\bullet,\bullet}$ is needed to get exactness of the full sequence. $\endgroup$ Sep 15 at 11:47
  • $\begingroup$ Thanks, I think the answer is totally clear now! $\endgroup$
    – Tom
    Sep 15 at 11:57

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