[proofplan] We define the two operators locally in holomorphic coordinates by applying the ordinary [exterior derivative](/theorems/1525) to the coefficient functions and separating the resulting terms according to whether they contain $dz_i$ or $d\bar z_i$. Holomorphic transition maps preserve the span of the $dz_i$ and separately preserve the span of the $d\bar z_i$, so these local definitions are independent of coordinates and glue to global operators. Finally, the identity $d^2=0$ decomposes by bidegree into the three identities for $\partial$ and $\bar{\partial}$. [/proofplan]

[step:Split the exterior derivative in holomorphic coordinates]Let $(U,z_1,\dots,z_n)$ be a holomorphic coordinate chart on $X$. For each $i \in \{1,\dots,n\}$, regard \begin{align*} z_i: U \to \mathbb{C} \end{align*} as a smooth complex-valued function, and write $dz_i$ and $d\bar z_i$ for the exterior derivatives of $z_i$ and its complex conjugate $\bar z_i$. Let $\alpha \in A^{p,q}(X)$. On $U$, every such form has a unique expression \begin{align*} \alpha|_U = \sum_{I,J} a_{I,J}\, dz_{i_1}\wedge \cdots \wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}, \end{align*} where $I=(i_1<\cdots<i_p)$ and $J=(j_1<\cdots<j_q)$ range over increasing index sets, and each coefficient is a smooth map \begin{align*} a_{I,J}: U \to \mathbb{C}. \end{align*} For a smooth function $f:U\to\mathbb{C}$, define its local $(1,0)$ and $(0,1)$ differentials in this chart by \begin{align*} \partial_U f = \sum_{k=1}^n \frac{\partial f}{\partial z_k}\, dz_k. \end{align*} and \begin{align*} \bar\partial_U f = \sum_{k=1}^n \frac{\partial f}{\partial \bar z_k}\, d\bar z_k. \end{align*} The exterior derivative of $f$ is then \begin{align*} df = \partial_U f + \bar\partial_U f. \end{align*} Since $d(dz_i)=d(d\bar z_i)=0$, the graded Leibniz rule for $d$ gives \begin{align*} d(\alpha|_U)=\sum_{I,J} da_{I,J}\wedge dz_{i_1}\wedge \cdots \wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}. \end{align*} Substituting $da_{I,J}=\partial_U a_{I,J}+\bar\partial_U a_{I,J}$ separates this expression into a component of type $(p+1,q)$ and a component of type $(p,q+1)$. Define \begin{align*} (\partial\alpha)|_U=\sum_{I,J} \partial_U a_{I,J}\wedge dz_{i_1}\wedge \cdots \wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}. \end{align*} and \begin{align*} (\bar\partial\alpha)|_U=\sum_{I,J} \bar\partial_U a_{I,J}\wedge dz_{i_1}\wedge \cdots \wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}. \end{align*} Thus, on $U$, \begin{align*} d(\alpha|_U)=(\partial\alpha)|_U+(\bar\partial\alpha)|_U, \end{align*} with $(\partial\alpha)|_U$ of type $(p+1,q)$ and $(\bar\partial\alpha)|_U$ of type $(p,q+1)$.[/step]

[guided]The local calculation begins with the defining feature of a complex coordinate chart: the one-forms $dz_1,\dots,dz_n,d\bar z_1,\dots,d\bar z_n$ form the natural complex coframe for complex-valued forms on $U$. Therefore a form $\alpha \in A^{p,q}(X)$ has, on $U$, a unique expansion \begin{align*} \alpha|_U = \sum_{I,J} a_{I,J}\, dz_{i_1}\wedge \cdots \wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}, \end{align*} where $I=(i_1<\cdots<i_p)$ and $J=(j_1<\cdots<j_q)$ are increasing index sets, and each coefficient is a smooth map \begin{align*} a_{I,J}: U \to \mathbb{C}. \end{align*} The key point is that $d$ differentiates only the coefficient functions, because the coordinate one-forms themselves are already exterior derivatives. For each $i$, the identity $d(dz_i)=0$ holds, and similarly $d(d\bar z_i)=0$. Applying the graded Leibniz rule to each summand gives \begin{align*} d(\alpha|_U)=\sum_{I,J} da_{I,J}\wedge dz_{i_1}\wedge \cdots \wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}. \end{align*} Now we split the ordinary differential of each coefficient according to complex type. For a smooth function $f:U\to\mathbb{C}$, define \begin{align*} \partial_U f = \sum_{k=1}^n \frac{\partial f}{\partial z_k}\, dz_k. \end{align*} and \begin{align*} \bar\partial_U f = \sum_{k=1}^n \frac{\partial f}{\partial \bar z_k}\, d\bar z_k. \end{align*} Then \begin{align*} df=\partial_U f+\bar\partial_U f. \end{align*} Substituting this into the formula for $d(\alpha|_U)$ produces two kinds of terms. The terms involving $\partial_U a_{I,J}$ add one $dz$ factor, so they have type $(p+1,q)$. The terms involving $\bar\partial_U a_{I,J}$ add one $d\bar z$ factor, so they have type $(p,q+1)$. This motivates the definitions \begin{align*} (\partial\alpha)|_U=\sum_{I,J} \partial_U a_{I,J}\wedge dz_{i_1}\wedge \cdots \wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}. \end{align*} and \begin{align*} (\bar\partial\alpha)|_U=\sum_{I,J} \bar\partial_U a_{I,J}\wedge dz_{i_1}\wedge \cdots \wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}. \end{align*} With these definitions, the local exterior derivative decomposes as \begin{align*} d(\alpha|_U)=(\partial\alpha)|_U+(\bar\partial\alpha)|_U. \end{align*} This is the desired decomposition locally: one part raises the holomorphic degree, and the other part raises the antiholomorphic degree.[/guided]

[step:Check that the local definitions are independent of holomorphic coordinates] Let $(V,w_1,\dots,w_n)$ be another holomorphic coordinate chart, and consider the overlap $U\cap V$. The coordinate transition map is holomorphic, so each function \begin{align*} w_a: U\cap V \to \mathbb{C} \end{align*} depends holomorphically on $z_1,\dots,z_n$. Hence \begin{align*} dw_a=\sum_{i=1}^n \frac{\partial w_a}{\partial z_i}\, dz_i. \end{align*} and \begin{align*} d\bar w_a=\sum_{i=1}^n \overline{\frac{\partial w_a}{\partial z_i}}\, d\bar z_i. \end{align*} Thus the transition from the $w$-coframe to the $z$-coframe sends the span of the $dw_a$ only into the span of the $dz_i$, and sends the span of the $d\bar w_a$ only into the span of the $d\bar z_i$. Therefore the decomposition of a form into its $(r,s)$-components is intrinsic on overlaps. Since $d\alpha$ is globally defined and its decomposition by type is unique on each overlap, the local $(p+1,q)$ component of $d\alpha$ computed in $z$-coordinates equals the one computed in $w$-coordinates, and similarly for the local $(p,q+1)$ component. Consequently the local definitions glue to global operators \begin{align*} \partial:A^{p,q}(X)\to A^{p+1,q}(X) \end{align*} and \begin{align*} \bar\partial:A^{p,q}(X)\to A^{p,q+1}(X). \end{align*} For every $\alpha\in A^{p,q}(X)$, the equality \begin{align*} d\alpha=\partial\alpha+\bar\partial\alpha \end{align*} holds globally. [/step]

[step:Extend the decomposition to all complex-valued forms] Let $A^k(X;\mathbb{C})$ denote the space of smooth complex-valued $k$-forms on $X$. The type decomposition gives \begin{align*} A^k(X;\mathbb{C})=\bigoplus_{p+q=k} A^{p,q}(X). \end{align*} For an arbitrary smooth complex-valued form $\omega\in A^k(X;\mathbb{C})$, write its unique decomposition as \begin{align*} \omega=\sum_{p+q=k}\omega_{p,q}, \end{align*} where $\omega_{p,q}\in A^{p,q}(X)$. Define \begin{align*} \partial\omega=\sum_{p+q=k}\partial\omega_{p,q}. \end{align*} and \begin{align*} \bar\partial\omega=\sum_{p+q=k}\bar\partial\omega_{p,q}. \end{align*} By the preceding step applied to each pure-type summand, \begin{align*} d\omega=\partial\omega+\bar\partial\omega. \end{align*} This proves the asserted decomposition of $d$ on all smooth complex-valued forms. [/step]

[step:Compare bidegrees in the identity $d^2=0$] Fix $\alpha\in A^{p,q}(X)$. Using $d=\partial+\bar\partial$, we compute \begin{align*} 0=d^2\alpha=(\partial+\bar\partial)(\partial\alpha+\bar\partial\alpha). \end{align*} Here we use the standard identity $d^2=0$ for the exterior derivative (citing a result not yet in the wiki: Exterior derivative squares to zero). Expanding the right-hand side gives \begin{align*} 0=\partial^2\alpha+\partial\bar\partial\alpha+\bar\partial\partial\alpha+\bar\partial^2\alpha. \end{align*} The four terms have distinct bidegrees except for the two middle terms: $\partial^2\alpha$ has type $(p+2,q)$, $\partial\bar\partial\alpha$ and $\bar\partial\partial\alpha$ have type $(p+1,q+1)$, and $\bar\partial^2\alpha$ has type $(p,q+2)$. Since the decomposition of a complex-valued form into bidegree components is direct, each bidegree component of the displayed sum must vanish. Therefore \begin{align*} \partial^2\alpha=0. \end{align*} and \begin{align*} \bar\partial^2\alpha=0. \end{align*} and \begin{align*} \partial\bar\partial\alpha+\bar\partial\partial\alpha=0. \end{align*} Because $\alpha\in A^{p,q}(X)$ was arbitrary and every complex-valued form is a finite sum of pure-type components, these identities hold as operator identities on smooth complex-valued forms. This completes the proof. [/step]