$\DeclareMathOperator{im}{im}$


I want to prove that Bott-Chern cohomology group $H^{p,q}\_{BC}=\frac{\ker\partial\cap \ker\bar{\partial}}{\im\partial\bar{\partial}}$ has finite dimension via Hodge decomposition like argument. References lead me to [this article](https://arxiv.org/abs/0709.3528) written by M. Schweitzer.


In 2.b he defines "Bott-Chern Laplacian" as $\Delta^{p,q}\_{BC}=(\partial\bar{\partial})(\partial\bar{\partial})^\*+\partial^\*\partial+\bar{\partial}^\*\bar{\partial}.$ He states that if $\Delta^{p,q}\_{BC}$ where elliptic, then we would have following decomposition:
$$\mathcal{E^{p,q}}=\mathcal{H}^{p,q}\_{\Delta\_{BC}}\oplus\im\partial\bar{\partial}\oplus (\im\partial^\*+\im\bar{\partial}^\*).\hspace{20pt}(\*)$$
So far so good. Cause $\Delta^{p,q}\_{BC}$ is self-adjoint and assuming that $\Delta^{p,q}\_{BC}$ is elliptic gives us that
$$\mathcal{E^{p,q}}=\mathcal{H}^{p,q}\_{\Delta\_{BC}}\oplus\im\Delta^{p,q}\_{BC}.$$
Repeating standard Hodge decomposition argument we notice that
$$\mathcal{H}^{p,q}\_{\Delta\_{BC}}\perp\im (\partial\bar{\partial})(\partial\bar{\partial})^\*\perp \im(\partial^\*\partial+\bar{\partial}^\*\bar{\partial})\perp\mathcal{H}^{p,q}\_{\Delta\_{BC}}$$
Additionally we easily verify that
$$\mathcal{H}^{p,q}\_{\Delta\_{BC}}\perp\im (\partial\bar{\partial})^\*\perp \im(\partial^\*\partial+\bar{\partial}^\*\bar{\partial})\hspace{5pt}\text{and} \hspace{5pt}\mathcal{H}^{p,q}\_{\Delta\_{BC}}\perp( \im\partial^\*+\im\bar{\partial}^\*)\perp \im (\partial\bar{\partial})(\partial\bar{\partial})^\*.$$
So $(\*)$ holds (assuming that $\Delta^{p,q}\_{BC}$ is elliptic).


Next he shows that $\Delta^{p,q}\_{BC}$ is not elliptic. As a result he introduces new operator
$$\tilde{\Delta}\_{BC}^{p,q}=(\partial\bar\partial)(\partial\bar\partial)^\*+(\partial\bar\partial)^\*(\partial\bar\partial)+(\bar\partial^\*\partial)(\bar\partial^\*\partial)^\*+(\bar\partial^\*\partial)^\*(\bar\partial^\*\partial)+\bar\partial^\*\bar\partial+\partial^\*\partial$$
and proves its ellipticity. After that he notice that
$$\tilde{\Delta}\_{BC}^{p,q}=0\iff \bar\partial=\partial=\bar\partial^\*\partial^\*=0.$$
Which in fact means that $\ker\Delta\_{BC}^{p,q}=\ker\tilde{\Delta}\_{BC}^{p,q},$ i.e. $\mathcal{H}^{p,q}\_{\Delta\_{BC}}=\mathcal{H}^{p,q}\_{\tilde{\Delta}\_{BC}}.$ As a result of the above he states Theorem 2.2 which claims exactly $(\*).$


I do not see how to infer $(\*)$. Since $\Delta\_{BC}^{p,q}$ is not elliptic, I believe we cannot infer $(\*)$ just by the fact that $\mathcal{H}^{p,q}\_{\Delta\_{BC}}=\mathcal{H}^{p,q}\_{\tilde{\Delta}\_{BC}}.$


PS. Article is written in French, so I might misunderstood something. Please correct me if I wrote something incorrectly.



Hodge decomposition for Bott-Chern cohomology
$\DeclareMathOperator{im}{im}$


Based on @YangMills suggestion I examined Bruno Bigolin's paper titled "Gruppi di Aeppli" and as a result I am now able to show the following relation



> 
> $$\ker\partial\cap\ker\bar\partial=\im\partial\bar\partial\oplus\_\perp\mathcal{H}.$$
> 
> 
> 


Let $\psi\in\ker\partial\cap\ker\bar\partial.$ We can represent $\psi$ as 
$$\psi=\Delta G\psi+\pi\psi=\\\color{red}{\partial\bar\partial\bar\partial^\*\partial^\* G\psi}+\color{blue}{\bar\partial^\*\partial^\*\partial\bar\partial G\psi}+\color{blue}{\bar\partial^\*\partial\partial^\*\bar\partial G\psi}+\color{green}{\partial^\*\bar\partial\bar\partial^\*\partial G\psi}+\color{blue}{\bar\partial^\*\bar\partial G\psi}+\color{green}{\partial^\*\partial G\psi} + \pi\psi=\\\color{red}{\partial\bar\partial\alpha}+\color{green}{\partial^\*\lambda}+\color{blue}{\bar\partial^\*\mu}+h,$$
where $G$ is Green's operator and $\pi$ is projection on $\mathcal{H}.$
As a result $$\partial^\*\lambda+\bar\partial^\*\mu=\psi-\partial\bar\partial\alpha-h.$$ Now since $\psi,\partial\bar\partial\alpha,h\in\ker\partial\cap\ker\bar\partial$, so as $\partial^\*\lambda+\bar\partial^\*\mu\in\ker\partial\cap\ker\bar\partial.$ And then
$$||\partial^\*\lambda+\bar\partial^\*\mu||^2=\langle\partial(\partial^\*\lambda+\bar\partial^\*\mu),\lambda\rangle+\langle\bar\partial(\partial^\*\lambda+\bar\partial^\*\mu),\mu\rangle=0.$$
Hence
$$\psi=\partial\bar\partial\alpha+h.$$
And yet $\im\partial\bar\partial\perp\mathcal{H}.$ In fact
$$\langle\partial\bar\partial\alpha,h\rangle=\langle\alpha,\bar\partial^\*\partial^\*h\rangle=0.$$
So we proved that
$$\ker\partial\cap\ker\bar\partial=\im\partial\bar\partial\oplus\_\perp\mathcal{H}.$$
Immediate consequence of above is that
$$H\_{BC}\cong\mathcal{H}.$$
And since $\Delta$ is elliptic we get that $\dim\_\mathbb{C}H\_{BC}<\infty.$


PS. Above $\Delta=\tilde\Delta\_{BC}^{p,q},\mathcal{H}=\mathcal{H}^{p,q}\_{\tilde\Delta\_{BC}}, H\_{BC}=H^{p,q}\_{BC}$ and $\psi\in\mathcal{E}^{p,q}.$


