# Hodge decomposition for Bott-Chern cohomology

$\DeclareMathOperator{im}{im}$

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 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.

• Yeah, he omitted some simple calculations. You can find the details for the dual case of Aeppli cohomology in an old paper of Bigolin archive.numdam.org/article/ASNSP_1969_3_23_2_259_0.pdf, see Theorem 1.7 and Corollary 1.8, pp.284-285. It is easy to adapt these to the case of Bott-Chern. Jun 21, 2016 at 15:19
• All you need to check (if you want, following Bigolin) is that the image of $\tilde{\Delta}_{BC}$ equals the image of $\partial\overline{\partial}$ plus the image of $\partial^*$ plus the image of $\overline{\partial}^*$. Jun 21, 2016 at 15:39
• see also the note on page 286 of Bigolin, where he remarks that the arguments he gives for Aeppli also apply to Bott-Chern, just switching $\partial$ with $\overline{\partial}^*$. Jun 21, 2016 at 16:24
• @YangMills This is exactly what I was looking for. I will go through the details and see if everything works flawless. Thank you. Jun 22, 2016 at 13:16

$\DeclareMathOperator{im}{im}$
$$\ker\partial\cap\ker\bar\partial=\im\partial\bar\partial\oplus_\perp\mathcal{H}.$$
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 $\psi\in\mathcal{E}^{p,q}.$