[proofplan] We prove the decomposition locally in holomorphic coordinates. A form of type $(p,q)$ is a sum of smooth coefficient functions times wedges of $dz_i$ and $d\bar z_j$, and differentiating such a term can only add one $dz_k$ or one $d\bar z_k$. This gives the two bidegree components of $d$, which are by definition $\partial$ and $\bar\partial$. Finally, applying the local coordinate formula twice and separating the resulting bidegrees gives $\partial^2=0$, $\bar\partial^2=0$, and $\partial\bar\partial+\bar\partial\partial=0$. [/proofplan] [step:Write a pure type form in holomorphic coordinates] Let $(U,\varphi)$ be a holomorphic coordinate chart on $M$, with coordinate functions \begin{align*} z_i:U\to\mathbb C \end{align*} for $1\le i\le n$. Write $z_i=x_i+i y_i$, where \begin{align*} x_i:U\to\mathbb R \end{align*} and \begin{align*} y_i:U\to\mathbb R \end{align*} are smooth real-valued functions. For an increasing multi-index $I=(i_1,\dots,i_p)$ with $1\le i_1<\dots<i_p\le n$, define \begin{align*} dz_I=dz_{i_1}\wedge\dots\wedge dz_{i_p}. \end{align*} For an increasing multi-index $J=(j_1,\dots,j_q)$ with $1\le j_1<\dots<j_q\le n$, define \begin{align*} d\bar z_J=d\bar z_{j_1}\wedge\dots\wedge d\bar z_{j_q}. \end{align*} On $U$, the [decomposition of complex differential forms by type](/theorems/7975) [citetheorem:7975] gives a unique expression \begin{align*} \alpha|_U=\sum_{\substack{|I|=p,\, |J|=q}} f_{I,J}\, dz_I\wedge d\bar z_J, \end{align*} where each coefficient is a smooth map \begin{align*} f_{I,J}:U\to\mathbb C. \end{align*} [guided] The point of passing to a holomorphic chart is that the local basis of complex-valued one-forms splits into the two types $dz_i$ and $d\bar z_i$. Since $\alpha$ has pure type $(p,q)$, the local decomposition by type [citetheorem:7975] says that on $U$ it is a sum of terms with exactly $p$ holomorphic one-form factors and exactly $q$ antiholomorphic one-form factors. More explicitly, for each increasing multi-index $I=(i_1,\dots,i_p)$ we define \begin{align*} dz_I=dz_{i_1}\wedge\dots\wedge dz_{i_p}. \end{align*} For each increasing multi-index $J=(j_1,\dots,j_q)$ we define \begin{align*} d\bar z_J=d\bar z_{j_1}\wedge\dots\wedge d\bar z_{j_q}. \end{align*} Then there are unique smooth coefficient functions \begin{align*} f_{I,J}:U\to\mathbb C \end{align*} such that \begin{align*} \alpha|_U=\sum_{\substack{|I|=p,\, |J|=q}} f_{I,J}\, dz_I\wedge d\bar z_J. \end{align*} This is the local form in which the [exterior derivative](/theorems/1525) can be inspected directly: all possible changes of type must come from differentiating the coefficient functions $f_{I,J}$, because the coordinate one-forms $dz_i$ and $d\bar z_i$ themselves are closed. [/guided] [/step] [step:Differentiate each local term and identify the two possible bidegrees] For each smooth coefficient map $f_{I,J}:U\to\mathbb C$, define the Wirtinger derivatives \begin{align*} \frac{\partial f_{I,J}}{\partial z_k}:U\to\mathbb C \end{align*} and \begin{align*} \frac{\partial f_{I,J}}{\partial \bar z_k}:U\to\mathbb C \end{align*} by \begin{align*} \frac{\partial f_{I,J}}{\partial z_k}=\frac{1}{2}\left(\frac{\partial f_{I,J}}{\partial x_k}-i\frac{\partial f_{I,J}}{\partial y_k}\right) \end{align*} and \begin{align*} \frac{\partial f_{I,J}}{\partial \bar z_k}=\frac{1}{2}\left(\frac{\partial f_{I,J}}{\partial x_k}+i\frac{\partial f_{I,J}}{\partial y_k}\right). \end{align*} The exterior derivative of the coefficient is \begin{align*} df_{I,J}=\sum_{k=1}^n \frac{\partial f_{I,J}}{\partial z_k}\,dz_k+\sum_{k=1}^n \frac{\partial f_{I,J}}{\partial \bar z_k}\,d\bar z_k. \end{align*} Since $d(dz_i)=0$ and $d(d\bar z_i)=0$ for every $1\le i\le n$, the graded Leibniz rule gives \begin{align*} d\left(f_{I,J}\,dz_I\wedge d\bar z_J\right)=df_{I,J}\wedge dz_I\wedge d\bar z_J. \end{align*} Substituting the displayed formula for $df_{I,J}$, we obtain \begin{align*} d\left(f_{I,J}\,dz_I\wedge d\bar z_J\right)=\sum_{k=1}^n \frac{\partial f_{I,J}}{\partial z_k}\,dz_k\wedge dz_I\wedge d\bar z_J+\sum_{k=1}^n \frac{\partial f_{I,J}}{\partial \bar z_k}\,d\bar z_k\wedge dz_I\wedge d\bar z_J. \end{align*} The first sum has type $(p+1,q)$, and the second sum has type $(p,q+1)$. Summing over all $I,J$ gives \begin{align*} d\alpha|_U\in\Omega^{p+1,q}(U)\oplus\Omega^{p,q+1}(U). \end{align*} Since this holds on every holomorphic coordinate chart, it holds globally on $M$. [/step] [step:Define the Dolbeault operators as the two bidegree projections of $d$] For $\alpha\in\Omega^{p,q}(M)$, define \begin{align*} \partial\alpha\in\Omega^{p+1,q}(M) \end{align*} to be the $(p+1,q)$ component of $d\alpha$, and define \begin{align*} \bar\partial\alpha\in\Omega^{p,q+1}(M) \end{align*} to be the $(p,q+1)$ component of $d\alpha$. These components are unique by the direct sum decomposition of forms by type [citetheorem:7975]. Therefore \begin{align*} d\alpha=\partial\alpha+\bar\partial\alpha \end{align*} for every pure type form $\alpha\in\Omega^{p,q}(M)$. Extending by linearity over the direct sum \begin{align*} \Omega^\bullet(M;\mathbb C)=\bigoplus_{0\le p,q\le n}\Omega^{p,q}(M) \end{align*} gives the operator identity \begin{align*} d=\partial+\bar\partial. \end{align*} The convention $\Omega^{a,b}(M)=\{0\}$ outside the range $0\le a,b\le n$ is consistent with the vanishing of forms outside the bidegree range [citetheorem:7977]. [/step] [step:Apply the decomposition twice and separate the bidegrees] Fix $\alpha\in\Omega^{p,q}(M)$. Using $d=\partial+\bar\partial$, compute \begin{align*} d^2\alpha=(\partial+\bar\partial)(\partial+\bar\partial)\alpha. \end{align*} Expanding by linearity gives \begin{align*} d^2\alpha=\partial^2\alpha+\partial\bar\partial\alpha+\bar\partial\partial\alpha+\bar\partial^2\alpha. \end{align*} The four terms have the following bidegrees: \begin{align*} \partial^2\alpha\in\Omega^{p+2,q}(M). \end{align*} \begin{align*} \partial\bar\partial\alpha+\bar\partial\partial\alpha\in\Omega^{p+1,q+1}(M). \end{align*} \begin{align*} \bar\partial^2\alpha\in\Omega^{p,q+2}(M). \end{align*} In local holomorphic coordinates, the identity $d^2=0$ follows from the displayed coordinate formula for $d$ and the equality of mixed second partial derivatives of smooth coefficient functions, together with antisymmetry of the wedge product. Hence \begin{align*} 0=d^2\alpha=\partial^2\alpha+\left(\partial\bar\partial\alpha+\bar\partial\partial\alpha\right)+\bar\partial^2\alpha. \end{align*} The three summands lie in distinct bidegree summands of the direct decomposition [citetheorem:7975]. Therefore uniqueness of the decomposition gives \begin{align*} \partial^2\alpha=0. \end{align*} \begin{align*} \partial\bar\partial\alpha+\bar\partial\partial\alpha=0. \end{align*} \begin{align*} \bar\partial^2\alpha=0. \end{align*} [guided] We now use the fact that the decomposition by type is a direct sum, so different bidegrees cannot cancel each other. Start with a fixed form $\alpha\in\Omega^{p,q}(M)$. Since the preceding step proved \begin{align*} d=\partial+\bar\partial, \end{align*} we may apply this identity twice: \begin{align*} d^2\alpha=(\partial+\bar\partial)(\partial+\bar\partial)\alpha. \end{align*} Expanding this expression by linearity gives \begin{align*} d^2\alpha=\partial^2\alpha+\partial\bar\partial\alpha+\bar\partial\partial\alpha+\bar\partial^2\alpha. \end{align*} The reason this expansion is useful is that each term has a prescribed bidegree. The operator $\partial$ increases the first bidegree by $1$ and leaves the second unchanged, while $\bar\partial$ leaves the first bidegree unchanged and increases the second by $1$. Therefore \begin{align*} \partial^2\alpha\in\Omega^{p+2,q}(M). \end{align*} Also both mixed terms have the same bidegree: \begin{align*} \partial\bar\partial\alpha\in\Omega^{p+1,q+1}(M) \end{align*} and \begin{align*} \bar\partial\partial\alpha\in\Omega^{p+1,q+1}(M). \end{align*} Finally, \begin{align*} \bar\partial^2\alpha\in\Omega^{p,q+2}(M). \end{align*} It remains to justify that the sum of these three bidegree components is zero. In the same local holomorphic coordinates used above, $d$ differentiates only the coefficient functions and wedges the result with $dz_k$ or $d\bar z_k$. Applying $d$ a second time produces second derivatives of the smooth coefficient functions. The terms with repeated coordinate directions vanish because a one-form wedged with itself is zero, and the terms with distinct coordinate directions cancel in pairs because mixed second partial derivatives of smooth functions agree while the wedge product changes sign when the two one-forms are interchanged. Hence $d^2\alpha=0$. Thus \begin{align*} 0=d^2\alpha=\partial^2\alpha+\left(\partial\bar\partial\alpha+\bar\partial\partial\alpha\right)+\bar\partial^2\alpha. \end{align*} The decomposition by type [citetheorem:7975] is direct, so a sum of forms in distinct bidegrees can be zero only if each bidegree component is zero. Consequently \begin{align*} \partial^2\alpha=0. \end{align*} \begin{align*} \partial\bar\partial\alpha+\bar\partial\partial\alpha=0. \end{align*} \begin{align*} \bar\partial^2\alpha=0. \end{align*} [/guided] [/step] [step:Conclude the operator identities on the full bigraded algebra] The form $\alpha\in\Omega^{p,q}(M)$ was arbitrary, and the integers $p,q$ ranged over all bidegrees with the zero-space convention outside $0\le p,q\le n$. Therefore the identities hold on every summand $\Omega^{p,q}(M)$ and hence, by linearity, on the full bigraded algebra of complex-valued differential forms: \begin{align*} \partial^2=0,\qquad \bar\partial^2=0,\qquad \partial\bar\partial+\bar\partial\partial=0. \end{align*} Together with \begin{align*} d=\partial+\bar\partial, \end{align*} this proves the claimed Dolbeault decomposition of the exterior derivative. [/step]