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Choose a positive Hermitian metric on $L$. Its curvature form $\omega =i{F}_h$ is a Kähler form on the compact complex manifold of dimension one. Part (c) gives a threshold $k_0$ with $H^1(X,\mathcal{O}(L^k))=0$ for every $k\ge k_0$. We use this to prescribe a finite principal part of a meromorphic section.

Fix a smooth cutoff $\chi$ supported inside the given coordinate chart and equal to one near $x_0$. Let $P(z)={\sum }_{j=-r}^{-1}{a}_j{z}^j$. On the punctured manifold define $t_k=\chi P(z){\zeta }^k$ inside the chart, extended by zero outside. Its Dolbeault operatoris smooth globally: it vanishes near the pole and near the boundary of the chart. It is ${\bar{\partial }}_{L^k}$-closed, either by the square-zero identity or because there are no $(0,2)$-forms in complex dimension one. The vanishing of $H^1(X,\mathcal{O}(L^k))$, together with the Dolbeault theorem, therefore gives a global smooth section $u_k$ with ${\bar{\partial }}_{L^k}{u}_k={\eta }_k$.

The section $s_k=t_k-u_k$ is holomorphic on $X\setminus \{x_0\}$. Near $x_0$, ${\eta }_k=0$, so $u_k=f_k(z){\zeta }^k$ with $f_k$ holomorphic through zero. Its Taylor series then giveswhere the second series converges near zero and its coefficients are those of $-f_k$. The threshold is determined by the fixed bundle $L$ and $X$, and is independent of the prescribed coefficients. In fact the same threshold works for every finite pole order $r$, since the Dolbeault cohomology obstruction is the same $H^1(X,L^k)$ for every such cutoff construction.

Choose a constant positive Hermitian form on $V$. Its underlying flat Kähler metric is translation-invariant and descends through the Euclidean lattice to the compact complex torus $X$. The Dolbeault Hodge decomposition gives a unique Dolbeault Laplacian-harmonic representative of every Dolbeault cohomology class. By the Kähler Laplacian identity, these are also Hodge Laplacian-harmonic.

Choose a constant coframe on $V$. A lifted harmonic form is a sum of constant wedge basis forms with smooth periodic coefficients. For the flat metric, the Hodge Laplacian acts coefficientwise by the scalar constant-coefficient Laplacian. Each coefficient is a harmonic function on a compact flat real torus, hence is constant: integration by parts gives $\int |Da{|}^2=0$. Conversely every constant-coefficient form is harmonic. Thus the harmonic representatives are exactly the translation-invariant differential forms of type (p, q).

The constant $(1,0)$ covectors form $V^{\ast }={\mathrm{Hom}}_{\mathbb{C}}(V,\mathbb{C})$; the constant $(0,1)$ covectors form the space of antilinear maps $\overline{V^{\ast }}={\mathrm{Hom}}_{\overline{\mathbb{C}}}(V,\mathbb{C})$. Sending a constant wedge form to its cohomology class givesThe map is intrinsic and does not depend on the auxiliary flat metric: that metric proves every class has exactly one constant representative, while the inclusion of constant forms is defined directly by the quotient. The Dolbeault cohomology of a complex torus consists of constant forms, and in particularFor indices outside $0,\dots ,g$, both the exterior-power expression and the cohomology group vanish.

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:

Let ${\Omega }^p$ be the sheaf of holomorphic $p$-forms and ${\mathcal{A}}^{p,q}$ the sheaf of smooth $(p,q)$-forms. The Dolbeault-Poincaré lemma makesan exact resolution by fine sheaves, which are acyclic for global sections. Its global cochain complex computes Dolbeault cohomology, proving the Dolbeault theorem

The Dolbeault cohomology group isThe complex orientation of the codimension-$r$ submanifold $Z$ gives an integral Poincaré-dual class $[Z]\in H^{2r}(X;\mathbb{Z})$. Integration over $Z$ defines a closed current of type $(r,r)$, so under the Dolbeault identification its complexification belongs to $H_{\bar{\partial }}^{r,r}(X)$.

A polydisc is a product of open discs in $\mathbb{C}$. It is a contractible complex manifold, with vanishing positive-degree de Rham cohomology and vanishing Dolbeault cohomology in positive antiholomorphic degree.

For a compact complex $n$-manifold and a holomorphic vector bundle $E$, analytic Serre duality givesThe pairing integrates the wedge product of bundle-valued Dolbeault cohomology representatives, contracting $E$ with its dual.