[proofplan] We work in a holomorphic coordinate polydisc around the chosen point and write the closed real $(1,1)$-form by its Hermitian coefficient functions. The equation $d\omega=0$, together with the type decomposition $d=\partial+\bar\partial$, gives the integrability relations needed to solve first a $\bar\partial$-system for a local $(1,0)$-form and then a $\partial$-system for a local complex-valued potential. Finally, the reality of $\omega$ lets us replace the complex potential by its real part, and positivity of $\omega$ is exactly positive definiteness of the resulting complex Hessian. [/proofplan]

[step:Choose holomorphic coordinates and record the coefficient form] Fix $p\in X$. Choose a holomorphic coordinate chart $(V,z)$ with $p\in V$, where \begin{align*} z:V\to z(V)\subset \mathbb C^n \end{align*} is biholomorphic onto an [open set](/page/Open%20Set). Shrinking $V$ if necessary, choose a coordinate polydisc $U\subset V$ with $p\in U$ and $z(U)$ convex. On $U$, write the smooth real $(1,1)$-form $\omega$ as \begin{align*} \omega=i\sum_{a=1}^n\sum_{b=1}^n h_{a\bar b}\,dz_a\wedge d\bar z_b, \end{align*} where each coefficient is a smooth function \begin{align*} h_{a\bar b}:U\to \mathbb C. \end{align*} Because $\omega$ is real, the coefficient matrix is Hermitian: \begin{align*} h_{a\bar b}=\overline{h_{b\bar a}} \end{align*} for all $1\le a,b\le n$. [/step]

[step:Extract the integrability relations from $d\omega=0$]By the [type decomposition of the exterior derivative](/theorems/7004) [citetheorem:7004], $d=\partial+\bar\partial$ on complex-valued forms. Since $\omega$ has type $(1,1)$, the equality $d\omega=0$ implies separately \begin{align*} \partial\omega=0 \end{align*} and \begin{align*} \bar\partial\omega=0. \end{align*} Computing the $(2,1)$-part gives, for all $1\le a,k,b\le n$, \begin{align*} \partial_{z_k}h_{a\bar b}=\partial_{z_a}h_{k\bar b}. \end{align*} Computing the $(1,2)$-part gives, for all $1\le a,b,k\le n$, \begin{align*} \partial_{\bar z_k}h_{a\bar b}=\partial_{\bar z_b}h_{a\bar k}. \end{align*}[/step]

[guided]The point of this step is to translate closedness of the differential form into compatibility equations for its coefficient functions. Since $\omega$ has type $(1,1)$, the form $\partial\omega$ has type $(2,1)$ and the form $\bar\partial\omega$ has type $(1,2)$. These two types live in different summands of the type decomposition of forms, so $d\omega=0$ forces both pieces to vanish. Using the coordinate expression \begin{align*} \omega=i\sum_{a=1}^n\sum_{b=1}^n h_{a\bar b}\,dz_a\wedge d\bar z_b, \end{align*} we compute \begin{align*} \partial\omega=i\sum_{k=1}^n\sum_{a=1}^n\sum_{b=1}^n \partial_{z_k}h_{a\bar b}\,dz_k\wedge dz_a\wedge d\bar z_b. \end{align*} The coefficient of $dz_k\wedge dz_a\wedge d\bar z_b$ must be skew-symmetric in $k$ and $a$. Therefore the vanishing of $\partial\omega$ is equivalent to \begin{align*} \partial_{z_k}h_{a\bar b}-\partial_{z_a}h_{k\bar b}=0. \end{align*} Hence \begin{align*} \partial_{z_k}h_{a\bar b}=\partial_{z_a}h_{k\bar b}. \end{align*} Similarly, \begin{align*} \bar\partial\omega=i\sum_{a=1}^n\sum_{b=1}^n\sum_{k=1}^n \partial_{\bar z_k}h_{a\bar b}\,d\bar z_k\wedge dz_a\wedge d\bar z_b. \end{align*} Moving $d\bar z_k$ past $dz_a$ introduces one minus sign: \begin{align*} d\bar z_k\wedge dz_a\wedge d\bar z_b=-dz_a\wedge d\bar z_k\wedge d\bar z_b. \end{align*} Thus the vanishing of the $(1,2)$-form is equivalent to the skew-[symmetry condition](/theorems/1360) in the barred indices: \begin{align*} \partial_{\bar z_k}h_{a\bar b}-\partial_{\bar z_b}h_{a\bar k}=0. \end{align*} So \begin{align*} \partial_{\bar z_k}h_{a\bar b}=\partial_{\bar z_b}h_{a\bar k}. \end{align*}[/guided]

[step:Solve the barred equations to build a $\partial$-closed $(1,0)$-form] For each fixed $a$, define the smooth $(0,1)$-form \begin{align*} \eta_a=\sum_{b=1}^n h_{a\bar b}\,d\bar z_b\in A^{0,1}(U). \end{align*} The relations \begin{align*} \partial_{\bar z_k}h_{a\bar b}=\partial_{\bar z_b}h_{a\bar k} \end{align*} say exactly that $\bar\partial\eta_a=0$. Since $z(U)$ is a convex polydisc, the local $\bar\partial$-Poincaré lemma applies to each $\eta_a$. Hence there are smooth functions \begin{align*} a_a:U\to\mathbb C \end{align*} such that \begin{align*} \bar\partial a_a=\eta_a. \end{align*} Equivalently, \begin{align*} \partial_{\bar z_b}a_a=h_{a\bar b} \end{align*} for all $a,b$. Define the smooth $(1,0)$-form \begin{align*} \alpha=\sum_{a=1}^n a_a\,dz_a\in A^{1,0}(U). \end{align*} Then \begin{align*} \bar\partial\alpha=\sum_{b=1}^n\sum_{a=1}^n h_{a\bar b}\,d\bar z_b\wedge dz_a. \end{align*} Since $d\bar z_b\wedge dz_a=-dz_a\wedge d\bar z_b$, this becomes \begin{align*} \bar\partial\alpha=-\sum_{a=1}^n\sum_{b=1}^n h_{a\bar b}\,dz_a\wedge d\bar z_b. \end{align*} Moreover, the remaining integrability relations allow us to modify $\alpha$ by a holomorphic $(1,0)$-form so that $\partial\alpha=0$ while preserving $\bar\partial\alpha$. Indeed, for every $1\le a,k,b\le n$, \begin{align*} \partial_{\bar z_b}\left(\partial_{z_k}a_a-\partial_{z_a}a_k\right) = \partial_{z_k}h_{a\bar b}-\partial_{z_a}h_{k\bar b} = 0. \end{align*} Thus each coefficient $\partial_{z_k}a_a-\partial_{z_a}a_k$ is holomorphic on $U$. Define the holomorphic $(2,0)$-form \begin{align*} \gamma:=\partial\alpha\in A^{2,0}(U). \end{align*} The Dolbeault operator relation $\partial^2=0$ [citetheorem:8046] gives \begin{align*} \partial\gamma=\partial^2\alpha=0. \end{align*} Since $z(U)$ is a convex polydisc, the holomorphic Poincaré lemma applies to the holomorphic closed $(2,0)$-form $\gamma$. Hence there is a holomorphic $(1,0)$-form \begin{align*} \beta=\sum_{a=1}^n b_a\,dz_a\in A^{1,0}(U), \end{align*} where each coefficient is a [holomorphic function](/page/Holomorphic%20Function) $b_a:U\to\mathbb C$, such that \begin{align*} \partial\beta=\gamma. \end{align*} Replace $\alpha$ by \begin{align*} \widetilde\alpha:=\alpha-\beta=\sum_{a=1}^n \widetilde a_a\,dz_a, \end{align*} where $\widetilde a_a:=a_a-b_a$ is a smooth function $\widetilde a_a:U\to\mathbb C$. Because each $b_a$ is holomorphic, $\bar\partial\beta=0$, so \begin{align*} \bar\partial\widetilde\alpha=\bar\partial\alpha. \end{align*} Thus the identities $\partial_{\bar z_b}\widetilde a_a=h_{a\bar b}$ are preserved. Also, \begin{align*} \partial\widetilde\alpha=\partial\alpha-\partial\beta=\gamma-\gamma=0. \end{align*} Renaming $\widetilde\alpha$ and $\widetilde a_a$ as $\alpha$ and $a_a$, respectively, we have chosen the functions $a_a$ so that \begin{align*} \partial\alpha=0. \end{align*} [/step]

[step:Integrate the $\partial$-closed form to a complex local potential] Since $\alpha\in A^{1,0}(U)$ is smooth and $\partial\alpha=0$, the local $\partial$-Poincaré lemma on the coordinate polydisc gives a smooth complex-valued function \begin{align*} f:U\to\mathbb C \end{align*} such that \begin{align*} \partial f=\alpha. \end{align*} Using the [Dolbeault operator relations](/theorems/8046) $\partial^2=0$ and $\partial\bar\partial+\bar\partial\partial=0$ [citetheorem:8046], we get \begin{align*} \partial\bar\partial f=-\bar\partial\partial f=-\bar\partial\alpha. \end{align*} From the computation of $\bar\partial\alpha$, \begin{align*} \partial\bar\partial f=\sum_{a=1}^n\sum_{b=1}^n h_{a\bar b}\,dz_a\wedge d\bar z_b. \end{align*} Multiplying by $i$ gives \begin{align*} i\partial\bar\partial f=\omega|_U. \end{align*} [/step]

[step:Replace the complex potential by a real-valued potential] Because $\omega$ is real, \begin{align*} \overline{\omega|_U}=\omega|_U. \end{align*} Taking complex conjugates in \begin{align*} \omega|_U=i\partial\bar\partial f \end{align*} and using the standard conjugation rule for type operators gives \begin{align*} \omega|_U=i\partial\bar\partial\overline{f}. \end{align*} Define the smooth real-valued function \begin{align*} \varphi:U\to\mathbb R,\qquad \varphi=\frac{f+\overline f}{2}. \end{align*} By complex linearity of $\partial\bar\partial$, \begin{align*} i\partial\bar\partial\varphi = \frac{1}{2}i\partial\bar\partial f+\frac{1}{2}i\partial\bar\partial\overline f = \omega|_U. \end{align*} Thus $\varphi$ is the required local real potential. [/step]

[step:Identify positivity with strict plurisubharmonicity of the potential] Assume now that $\omega$ is positive definite as a real $(1,1)$-form. Since \begin{align*} i\partial\bar\partial\varphi = i\sum_{a=1}^n\sum_{b=1}^n \partial_{z_a}\partial_{\bar z_b}\varphi\,dz_a\wedge d\bar z_b, \end{align*} comparison with the coefficient expression for $\omega$ gives \begin{align*} h_{a\bar b}=\partial_{z_a}\partial_{\bar z_b}\varphi. \end{align*} For every point $q\in U$, positivity of $\omega_q$ means that the Hermitian matrix \begin{align*} \left(h_{a\bar b}(q)\right)_{1\le a,b\le n} \end{align*} is positive definite. Hence the complex Hessian matrix \begin{align*} \left(\partial_{z_a}\partial_{\bar z_b}\varphi(q)\right)_{1\le a,b\le n} \end{align*} is positive definite for every $q\in U$. This is precisely the coordinate criterion for $\varphi$ to be strictly plurisubharmonic on $U$. Therefore, when $\omega$ is positive definite, the local potential $\varphi$ is strictly plurisubharmonic. [/step]