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For a differential form of type (p, q), $d\alpha =0$ is equivalent to both $\partial \alpha =0$ and $\bar{\partial }\alpha =0$, because the two resulting types are distinct. Moreover $\partial \bar{\partial }\beta$ has type $(p,q)$ when $\beta$ has type $(p-1,q-1)$, andby the square-zero and anticommutation identities. Thus the denominator defining Bott-Chern cohomology is a vector subspace of its numerator, making the quotient well defined.

Complex conjugation of differential-form type sends a $d$-closed $(p,q)$-form to a $d$-closed $(q,p)$-form. For a representative in the denominator,The minus sign does not change the denominator vector subspace. Conjugation consequently induces a conjugate-linear bijection $[\alpha ]\mapsto [\bar{\alpha }]$ between the two Bott-Chern cohomology spaces; applying it twice is the identity. Equivalently, $\boxed{H_{BC}^{p,q}(X)\cong \overline{H_{BC}^{q,p}(X)}}$ as complex vector spaces.

A $d$-closed differential form of type (p, q) is $\bar{\partial }$-closed, and a Bott-Chern cohomology coboundary is a Dolbeault cohomology coboundary becauseHence taking the same representative defines a complex-linear map $\varphi :H_{BC}^{p,q}(X)\to H_{\bar{\partial }}^{p,q}(X)$.

If $X$ is compact and Kähler, the Dolbeault Hodge decomposition supplies a $\bar{\partial }$-harmonic representative $\alpha$ of every Dolbeault cohomology class. The scalar Kähler Laplacian identity is ${\Delta }_d=2{\Delta }_{\bar{\partial }}$, so $\alpha$ is also $d$-harmonic. Consequentlyand $d\alpha =0$. It therefore defines a Bott-Chern cohomology class mapping to the original class. Thus $\boxed{\varphi \text{ is surjective on a compact Ka¨hler manifold}.}$

For clarity, the scalar Kähler Laplacian identity follows from the Kähler identities: they make the mixed anticommutators $\{\partial ,{\bar{\partial }}^{\ast }\}$ and $\{\bar{\partial },{\partial }^{\ast }\}$ zero, while the curvature-zero instance of the identity proved in 4(b) gives ${\Delta }^{\prime }={\Delta }^{\prime \prime }$. Expanding ${\Delta }_d$ yields ${\Delta }^{\prime }+{\Delta }^{\prime \prime }=2{\Delta }^{\prime \prime }$.

The natural map between Bott-Chern cohomology and Dolbeault cohomology isIt is well-defined: a $d$-closed pure-type form is $\bar{\partial }$-closed, and $\partial \bar{\partial }\beta =-\bar{\partial }(\partial \beta )$ is $\bar{\partial }$-exact.

For surjectivity, choose the unique harmonic differential form $h$ representing a Dolbeault cohomology class using Dolbeault Hodge decomposition. The Kähler Laplacian identity ${\Delta }_d=2{\Delta }_{\bar{\partial }}$ makes $h$ a harmonic differential form for $d$, hence $d$-closed. Sending the Dolbeault class to $[h{]}_{BC}$ consequently gives a right inverse.

For injectivity, the ddbar lemma says directly that a $d$-closed, $\bar{\partial }$-exact pure-type form is $\partial \bar{\partial }$-exact. One can also exhibit its primitive. Let $G_d$ be the Green operator of the Hodge Laplacian, and put $G_{\bar{\partial }}=2G_d$. If $\alpha$ is $d$-closed and $\bar{\partial }$-exact, its harmonic projection vanishes and the Kähler identities giveHere $\partial G_{\bar{\partial }}\alpha =\bar{\partial }{G}_{\bar{\partial }}\alpha =0$ because the Green operator commutes with these differentials, and ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. Thus $\alpha$ represents zero in Bott-Chern cohomology. We have constructed the canonical isomorphism and its harmonic inverse: