# The kernel of $H^{\bullet,\bullet}_{\bar\partial}(X)\to H_A^{\bullet,\bullet}(X)$

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 $$B^{\bullet,\bullet}=\frac{\ker\bar\partial\cap\text{im }\partial}{\text{im }\partial\bar\partial}$$?

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 $$\operatorname{im}f \subseteq \ker g$$.

• The explanation is precise! But why the denominator of $B^{\bullet,\bullet}$ should be $\text{im }\partial\bar\partial$?
– Tom
Sep 15 at 3:07
• 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. Sep 15 at 11:47
• Thanks, I think the answer is totally clear now!
– Tom
Sep 15 at 11:57