In order to establish the Dolbeault isomorphism we need to prove the
Dolbeault–Grothendieck lemma (or
\bar{\partial}-Poincaré lemma). First we prove a one-dimensional version of the \bar{\partial}-Poincaré lemma; we shall use the following generalised form of the
Cauchy integral representation for smooth functions:
Proposition: Let B_{\varepsilon}(0):=\lbrace z\in\Complex \mid | z| the open ball centered in 0 of radius \varepsilon\in\R _{>0}, \overline{B_{\varepsilon}(0)}\subseteq U open and f\in\mathcal{C}^\infty(U), then :\forall z\in B_{\varepsilon}(0): \quad f(z)=\frac{1}{2\pi i}\int_{\partial B_{\varepsilon}(0)}\frac{f(\xi)}{\xi-z}d\xi+\frac{1}{2\pi i}\iint_{B_{\varepsilon}(0)}\frac{\partial f}{\partial\bar{\xi}}\frac{d\xi\wedge d\bar{\xi}}{\xi-z}.
Lemma (\bar{\partial}-Poincaré lemma on the
complex plane): Let B_{\varepsilon}(0),U be as before and \alpha=f d\bar{z}\in\mathcal{A}^{0,1}_{\Complex}(U) a smooth form, then :\mathcal{C}^\infty(U)\ni g(z):=\frac{1}{2\pi i}\int_{B_{\varepsilon}(0)}\frac{f(\xi)}{\xi-z}d\xi\wedge d\bar{\xi} satisfies \alpha=\bar{\partial}g on B_{\varepsilon}(0).
Proof. Our claim is that g defined above is a well-defined smooth function and \alpha = f\, d\bar{z} = \bar{\partial} g. To show this we choose a point z\in B_{\varepsilon}(0) and an open neighbourhood z\in V\subseteq B_{\varepsilon}(0), then we can find a smooth function \rho: B_{\varepsilon}(0)\to\R whose support is compact and lies in B_{\varepsilon}(0) and \rho|_V\equiv 1. Then we can write :f=f_1+f_2:=\rho f+(1-\rho)f and define : g_i:=\frac{1}{2\pi i}\int_{B_{\varepsilon}(0)}\frac{f_i(\xi)}{\xi-z}d\xi\wedge d\bar{\xi}. Since f_2\equiv 0 in V then g_2 is clearly well-defined and smooth; we note that :\begin{align} g_1&=\frac{1}{2\pi i}\int_{B_{\varepsilon}(0)}\frac{f_1(\xi)}{\xi-z}d\xi\wedge d\bar{\xi}\\ &=\frac{1}{2\pi i}\int_{\Complex}\frac{f_1(\xi)}{\xi-z}d\xi\wedge d\bar{\xi}\\ &=\pi^{-1}\int_0^\infty\int_0^{2\pi}f_1(z+re^{i\theta})e^{-i\theta}d\theta dr, \end{align} which is indeed well-defined and smooth, therefore the same is true for g. Now we show that \bar{\partial}g=\alpha on B_{\varepsilon}(0). :\frac{\partial g_2}{\partial\bar{z}}=\frac{1}{2\pi i}\int_{B_{\varepsilon}(0)}f_2(\xi)\frac{\partial}{\partial\bar{z}}\Big(\frac{1}{\xi-z}\Big)d\xi\wedge d\bar{\xi}=0 since (\xi-z)^{-1} is holomorphic in B_{\varepsilon}(0)\setminus V . :\begin{align} \frac{\partial g_1}{\partial \bar{z}}=&\pi^{-1}\int_{\Complex}\frac{\partial f_1(z+re^{i\theta})}{\partial\bar{z}} e^{-i\theta}d\theta\wedge dr\\ =& \pi^{-1}\int_{\Complex }\Big(\frac{\partial f_1}{\partial\bar{z}}\Big)(z+re^{i\theta}) e^{-i\theta}d\theta\wedge dr\\ =&\frac{1}{2\pi i}\iint_{B_{\varepsilon}(0)}\frac{\partial f_1}{\partial\bar{\xi}}\frac{d\xi\wedge d\bar{\xi}}{\xi-z} \end{align} applying the generalised Cauchy formula to f_1 we find :f_1(z)=\frac{1}{2\pi i}\int_{\partial B_{\varepsilon}(0)}\frac{f_1(\xi)}{\xi-z}d\xi+\frac{1}{2\pi i}\iint_{B_{\varepsilon}(0)}\frac{\partial f_1}{\partial\bar{\xi}}\frac{d\xi\wedge d\bar{\xi}}{\xi-z} =\frac{1}{2\pi i}\iint_{B_{\varepsilon}(0)}\frac{\partial f_1}{\partial\bar{\xi}}\frac{d\xi\wedge d\bar{\xi}}{\xi-z} since f_1|_{\partial B_{\varepsilon}(0)}=0, but then f=f_1=\frac{\partial g_1}{\partial\bar{z}}=\frac{\partial g}{\partial\bar{z}} on V. Since z was arbitrary, the lemma is now proved.
Proof of Dolbeault–Grothendieck lemma Now are ready to prove the Dolbeault–Grothendieck lemma; the proof presented here is due to
Grothendieck. We denote with \Delta_{\varepsilon}^n(0) the open
polydisc centered in 0\in\Complex^n with radius \varepsilon\in\R_{>0}.
Lemma (Dolbeault–Grothendieck): Let \alpha\in\mathcal{A}_{\Complex^n}^{p,q}(U) where \overline{\Delta_{\varepsilon}^n(0)} \subseteq U open and q>0 such that \bar{\partial}\alpha=0, then there exists \beta\in\mathcal{A}_{\Complex^n}^{p,q-1}(U) which satisfies: \alpha=\bar{\partial}\beta on \Delta_{\varepsilon}^n(0). Before starting the proof we note that any (p,q)-form can be written as : \alpha=\sum_{IJ}\alpha_{IJ} dz_I\wedge d\bar{z}_J=\sum_{J}\left(\sum_I\alpha_{IJ} dz_I\right)_J\wedge d\bar{z}_J for multi-indices I,J,|I|=p,|J|=q, therefore we can reduce the proof to the case \alpha\in\mathcal{A}_{\Complex ^n}^{0,q}(U).
Proof. Let k>0 be the smallest index such that \alpha\in(d\bar{z}_1,\dots,d\bar{z}_k) in the sheaf of \mathcal{C}^\infty-modules, we proceed by induction on k. For k=0 we have \alpha\equiv0 since q>0; next we suppose that if \alpha\in(d\bar{z}_1,\dots,d\bar{z}_k) then there exists \beta\in\mathcal{A}_{\Complex^n}^{0,q-1}(U) such that \alpha=\bar{\partial}\beta on \Delta_{\varepsilon}^n(0). Then suppose \omega \in (d\bar{z}_1, \dots, d\bar{z}_{k+1}) and observe that we can write :\omega=d\bar{z}_{k+1}\wedge \psi + \mu, \qquad \psi, \mu\in(d\bar{z}_1,\dots,d\bar{z}_k). Since \omega is \bar{\partial}-closed it follows that \psi, \mu are holomorphic in variables z_{k+2},\dots,z_n and smooth in the remaining ones on the polydisc \Delta_{\varepsilon}^n(0). Moreover we can apply the \bar{\partial}-Poincaré lemma to the smooth functions z_{k+1} \mapsto \psi_J(z_1, \dots,z_{k+1}, \dots, z_n) on the open ball B_{\varepsilon_{k+1}}(0), hence there exist a family of smooth functions g_J which satisfy :\psi_J=\frac{\partial g_J}{\partial\bar{z}_{k+1}}\quad \text{on} \quad B_{\varepsilon_{k+1}}(0). g_J are also holomorphic in z_{k+2},\dots,z_n. Define :\tilde{\psi}:=\sum_J g_J d\bar{z}_{J} then :\begin{align} \omega-\bar{\partial}\tilde{\psi}&=d\bar{z}_{k+1}\wedge\psi +\mu-\sum_J\frac{\partial g_J}{\partial\bar{z}_{k+1}}d\bar{z}_{k+1}\wedge d\bar{z}_J +\sum_{j=1}^k\sum_J\frac{\partial g_J}{\partial \bar{z}_{j}}d\bar{z}_j\wedge d\bar{z}_{J\setminus\lbrace j\rbrace}\\ &=d\bar{z}_{k+1}\wedge\psi+\mu-d\bar{z}_{k+1}\wedge\psi+\sum_{j=1}^k\sum_J\frac{\partial g_J}{\partial \bar{z}_{j}}d\bar{z}_j\wedge d\bar{z}_{J\setminus\lbrace j\rbrace}\\ &=\mu+\sum_{j=1}^k\sum_J\frac{\partial g_J}{\partial\bar{z}_{j}}d\bar{z}_j\wedge d\bar{z}_{J\setminus\lbrace j \rbrace} \in (d\bar{z}_1, \dots, d\bar{z}_{k}), \end{align} therefore we can apply the induction hypothesis to it, there exists \eta\in\mathcal{A}_{\Complex ^n}^{0,q-1}(U) such that :\omega-\bar{\partial}\tilde{\psi}=\bar{\partial}\eta \quad \text{on} \quad \Delta_{\varepsilon}^n(0) and \zeta:=\eta+\tilde{\psi} ends the induction step. QED :The previous lemma can be generalised by admitting polydiscs with \varepsilon_k=+\infty for some of the components of the polyradius.
Lemma (extended Dolbeault-Grothendieck). If \Delta_\varepsilon^n(0) is an open polydisc with \varepsilon_k\in\R \cup \lbrace +\infty \rbrace and q>0, then H^{p,q}_{\bar{\partial}}(\Delta_\varepsilon^n(0))=0.
Proof. We consider two cases: \alpha\in\mathcal{A}_{\Complex^n}^{p,q+1}(U), q>0 and \alpha\in\mathcal{A}_{\Complex^n}^{p,1}(U). Case 1. Let \alpha\in\mathcal{A}_{\Complex^n}^{p,q+1}(U), q>0, and we cover \Delta_\varepsilon^n(0) with polydiscs \overline{\Delta_i} \subset\Delta_{i+1}, then by the Dolbeault–Grothendieck lemma we can find forms \beta_i of bidegree (p,q-1) on \overline{\Delta_i}\subseteq U_i open such that \alpha |_{\Delta_i} = \bar{\partial} \beta_i; we want to show that :\beta_{i+1}|_{\Delta_i}=\beta_i. We proceed by induction on i: the case when i=1 holds by the previous lemma. Let the claim be true for k>1 and take \Delta_{k+1} with :\Delta_{\varepsilon}^n(0)=\bigcup_{i=1}^{k+1}\Delta_i \quad \text{and} \quad \overline{\Delta_k}\subset\Delta_{k+1}. Then we find a (p,q-1)-form \beta'_{k+1} defined in an open neighbourhood of \overline{\Delta_{k+1}} such that \alpha|_{\Delta_{k+1}}=\bar{\partial}\beta_{k+1}. Let U_k be an open neighbourhood of \overline{\Delta_k} then \bar{\partial}(\beta_k-\beta'_{k+1})=0 on U_k and we can apply again the Dolbeault-Grothendieck lemma to find a (p,q-2)-form \gamma_k such that \beta_k-\beta'_{k+1}=\bar{\partial}\gamma_k on \Delta_k. Now, let V_k be an
open set with \overline{\Delta_k} \subset V_k \subsetneq U_k and \rho_k: \Delta_\varepsilon^n(0)\to\R a smooth function such that: :\operatorname{supp}(\rho_k)\subset U_k, \qquad \rho|_{V_k}=1, \qquad \rho_k|_{\Delta_\varepsilon^n(0)\setminus U_k}=0. Then \rho_k\gamma_k is a well-defined smooth form on \Delta_\varepsilon^n(0) which satisfies :\beta_k=\beta'_{k+1}+\bar{\partial}(\gamma_k\rho_k) \quad \text{on} \quad \Delta_{k}, hence the form :\beta_{k+1}:=\beta'_{k+1}+\bar{\partial}(\gamma_k\rho_k) satisfies :\begin{align} \beta_{k+1}|_{\Delta_k} &=\beta'_{k+1}+\bar{\partial}\gamma_k=\beta_k\\ \bar{\partial}\beta_{k+1}&=\bar{\partial}\beta'_{k+1}=\alpha|_{\Delta_{k+1}} \end{align} Case 2. If instead \alpha\in\mathcal{A}_{\Complex^n}^{p,1}(U), we cannot apply the Dolbeault-Grothendieck lemma twice; we take \beta_i and \Delta_i as before, we want to show that :\left \| \left. \left ({\beta_i}_I-{\beta_{i+1}}_I \right ) \right |_{\Delta_{k-1}} \right \|_\infty Again, we proceed by induction on i: for i=1 the answer is given by the Dolbeault-Grothendieck lemma. Next we suppose that the claim is true for k>1. We take \Delta_{k+1}\supset\overline{\Delta_k} such that \Delta_{k+1}\cup\lbrace\Delta_i\rbrace_{i=1}^k covers \Delta_\varepsilon^n(0), then we can find a (p,0)-form \beta'_{k+1} such that :\alpha|_{\Delta_{k+1}}=\bar{\partial}\beta'_{k+1}, which also satisfies \bar{\partial}(\beta_k-\beta'_{k+1})=0 on \Delta_k, i.e. \beta_k-\beta'_{k+1} is a holomorphic (p,0)-form wherever defined, hence by the
Stone–Weierstrass theorem we can write it as :\beta_k-\beta'_{k+1}=\sum_{|I|=p}(P_I+r_I)dz_I where P_I are polynomials and :\left \| r_I|_{\Delta_{k-1}} \right \|_\infty but then the form : \beta_{k+1}:=\beta'_{k+1}+\sum_{|I|=p}P_Idz_I satisfies :\begin{align} \bar{\partial}\beta_{k+1}&=\bar{\partial}\beta'_{k+1}=\alpha|_{\Delta_{k+1}}\\ \left \| ({\beta_k}_I-{\beta_{k+1}}_I)|_{\Delta_{k-1}} \right \|_\infty&=\| r_I\|_\infty which completes the induction step; therefore we have built a sequence \lbrace\beta_i\rbrace_{i\in\N} which uniformly converges to some (p,0)-form \beta such that \alpha|_{\Delta_{\varepsilon}^n(0)}=\bar{\partial}\beta. QED ==Dolbeault's theorem==