[proofplan] We prove $\bar\partial^2 = 0$ as an operator on smooth differential forms on an [open set](/page/Open%20Set) $\Omega \subset \mathbb{C}^n$. The proof has two parts: first we verify $\bar\partial^2 f = 0$ for smooth functions $f$ by computing the double application of $\bar\partial$ and using the symmetry of mixed partial derivatives against the antisymmetry of the wedge product. Then we extend to arbitrary $(p,q)$-forms by the Leibniz rule. [/proofplan]

[step:Compute $\bar\partial^2 f$ for a smooth function and show it vanishes by symmetry-antisymmetry cancellation]Let $f \in C^\infty(\Omega)$. By definition of $\bar\partial$: \begin{align*} \bar\partial f = \sum_{k=1}^n \frac{\partial f}{\partial\bar{z}_k}\,d\bar{z}_k. \end{align*} Applying $\bar\partial$ again to this $(0,1)$-form, using the Leibniz rule $\bar\partial(\alpha_k\,d\bar{z}_k) = (\bar\partial\alpha_k) \wedge d\bar{z}_k$ (since $d\bar{z}_k$ is a constant-coefficient form with $\bar\partial(d\bar{z}_k) = 0$): \begin{align*} \bar\partial^2 f = \bar\partial\!\left(\sum_{k=1}^n \frac{\partial f}{\partial\bar{z}_k}\,d\bar{z}_k\right) = \sum_{k=1}^n \left(\sum_{j=1}^n \frac{\partial^2 f}{\partial\bar{z}_j\,\partial\bar{z}_k}\,d\bar{z}_j\right) \wedge d\bar{z}_k = \sum_{j=1}^n \sum_{k=1}^n \frac{\partial^2 f}{\partial\bar{z}_j\,\partial\bar{z}_k}\,d\bar{z}_j \wedge d\bar{z}_k. \end{align*} We split this double sum into diagonal terms ($j = k$), terms with $j < k$, and terms with $j > k$. **Diagonal terms ($j = k$):** $d\bar{z}_j \wedge d\bar{z}_j = 0$ by the antisymmetry of the wedge product, so these terms vanish. **Off-diagonal terms:** For each pair $j < k$, the sum contains two terms: \begin{align*} \frac{\partial^2 f}{\partial\bar{z}_j\,\partial\bar{z}_k}\,d\bar{z}_j \wedge d\bar{z}_k + \frac{\partial^2 f}{\partial\bar{z}_k\,\partial\bar{z}_j}\,d\bar{z}_k \wedge d\bar{z}_j. \end{align*} Since $f \in C^\infty(\Omega)$, the mixed partial derivatives commute: $\frac{\partial^2 f}{\partial\bar{z}_j\,\partial\bar{z}_k} = \frac{\partial^2 f}{\partial\bar{z}_k\,\partial\bar{z}_j}$. Since the wedge product is antisymmetric: $d\bar{z}_k \wedge d\bar{z}_j = -d\bar{z}_j \wedge d\bar{z}_k$. Therefore the two terms cancel: \begin{align*} \frac{\partial^2 f}{\partial\bar{z}_j\,\partial\bar{z}_k}\,d\bar{z}_j \wedge d\bar{z}_k + \frac{\partial^2 f}{\partial\bar{z}_j\,\partial\bar{z}_k}\,(-d\bar{z}_j \wedge d\bar{z}_k) = 0. \end{align*} Since all diagonal and off-diagonal terms vanish, $\bar\partial^2 f = 0$.[/step]

[guided]The cancellation mechanism is the same one that makes $d^2 = 0$ in real differential geometry: symmetric coefficients paired with antisymmetric basis elements. Here, the symmetry of the coefficients $\frac{\partial^2 f}{\partial\bar{z}_j\,\partial\bar{z}_k}$ follows from the equality of mixed partials (Clairaut's theorem, valid for $C^2$ functions), while the antisymmetry $d\bar{z}_j \wedge d\bar{z}_k = -d\bar{z}_k \wedge d\bar{z}_j$ is a defining property of the exterior algebra. Why does this argument require smoothness? The equality of mixed partial derivatives $\frac{\partial^2 f}{\partial\bar{z}_j\,\partial\bar{z}_k} = \frac{\partial^2 f}{\partial\bar{z}_k\,\partial\bar{z}_j}$ holds for $f \in C^2$, but can fail for functions that are merely differentiable. For the $\bar\partial^2 = 0$ identity to hold in the distributional sense (as needed in Hörmander's $L^2$ theory), one must interpret $\bar\partial$ as a closed densely-defined operator and verify the domain conditions. Note that this computation also shows $\partial^2 = 0$ by the identical argument with $z_j$ replacing $\bar{z}_j$.[/guided]

[step:Extend to arbitrary $(p,q)$-forms by the Leibniz rule] A general smooth $(p,q)$-form on $\Omega$ can be written as \begin{align*} \alpha = \sum_{|J|=p,\,|K|=q} a_{JK}\,dz_J \wedge d\bar{z}_K, \end{align*} where $J = (j_1 < \dots < j_p)$ and $K = (k_1 < \dots < k_q)$ are increasing multi-indices, $a_{JK} \in C^\infty(\Omega)$, $dz_J = dz_{j_1} \wedge \dots \wedge dz_{j_p}$, and $d\bar{z}_K = d\bar{z}_{k_1} \wedge \dots \wedge d\bar{z}_{k_q}$. The $\bar\partial$ operator acts by the Leibniz rule: \begin{align*} \bar\partial\alpha = \sum_{J,K} (\bar\partial a_{JK}) \wedge dz_J \wedge d\bar{z}_K, \end{align*} where we use $\bar\partial(dz_J) = 0$ (since each $dz_{j_i}$ has $\bar\partial(dz_{j_i}) = 0$) and $\bar\partial(d\bar{z}_K) = 0$ (since each $d\bar{z}_{k_i}$ is a constant-coefficient $1$-form). Applying $\bar\partial$ a second time: \begin{align*} \bar\partial^2\alpha = \sum_{J,K} (\bar\partial^2 a_{JK}) \wedge dz_J \wedge d\bar{z}_K. \end{align*} Here we use the Leibniz rule for $\bar\partial$ applied to $(\bar\partial a_{JK}) \wedge dz_J \wedge d\bar{z}_K$: since $dz_J \wedge d\bar{z}_K$ is a constant-coefficient form, $\bar\partial((\bar\partial a_{JK}) \wedge dz_J \wedge d\bar{z}_K) = (\bar\partial^2 a_{JK}) \wedge dz_J \wedge d\bar{z}_K$. Since $a_{JK} \in C^\infty(\Omega)$, the previous step gives $\bar\partial^2 a_{JK} = 0$ for each coefficient. Therefore \begin{align*} \bar\partial^2\alpha = \sum_{J,K} 0 \cdot dz_J \wedge d\bar{z}_K = 0. \end{align*} This holds for every smooth $(p,q)$-form $\alpha$ on $\Omega$, completing the proof that $\bar\partial^2 = 0$ as an operator on $\mathcal{E}^{p,q}(\Omega)$ for all $p, q \geq 0$. [/step]