[proofplan] We prove the equivalence locally in holomorphic coordinates. The $(2,1)$-component $\partial\omega$ of $d\omega$ has coefficients given by the skew differences $\partial_{z_i}h_{j\bar k}-\partial_{z_j}h_{i\bar k}$. The Chern torsion has exactly the same skew differences after raising the barred index with the inverse Hermitian matrix $(h^{p\bar k})$. Since the metric matrix is invertible and $\omega$ is real, vanishing of the Chern torsion is equivalent to $\partial\omega=0$, and this is equivalent to $d\omega=0$. [/proofplan] [step:Write the Hermitian form and Chern torsion in holomorphic coordinates] Let $U \subset X$ be a coordinate neighbourhood with holomorphic coordinate map \begin{align*} z: U \to z(U) \subset \mathbb{C}^n. \end{align*} Write $z_i: U \to \mathbb{C}$ for the $i$-th coordinate function, $\partial_{z_i}$ for the corresponding local frame of $T^{1,0}X$, and $dz_i$ for the dual coframe. Define the local Hermitian metric coefficients \begin{align*} h_{i\bar j}: U \to \mathbb{C}, \qquad h_{i\bar j}=h(\partial_{z_i},\partial_{z_j}). \end{align*} Let $(h^{p\bar q})$ denote the inverse matrix to $(h_{i\bar j})$, so that \begin{align*} \sum_{q=1}^n h^{p\bar q}h_{r\bar q}=\delta_{pr}. \end{align*} With the standard convention for the associated fundamental form, \begin{align*} \omega = i\sum_{j,k=1}^n h_{j\bar k}\, dz_j \wedge d\bar z_k. \end{align*} For the Chern connection $\nabla^C$, define the local connection coefficients $\Gamma^p_{ij}: U \to \mathbb{C}$ by \begin{align*} \nabla^C_{\partial_{z_i}}\partial_{z_j}=\sum_{p=1}^n \Gamma^p_{ij}\partial_{z_p}. \end{align*} The defining [local formula for the Chern connection](/theorems/7049) of a Hermitian metric is \begin{align*} \Gamma^p_{ij}=\sum_{q=1}^n h^{p\bar q}\partial_{z_i}h_{j\bar q}. \end{align*} Define the Chern torsion coefficients $T^p_{ij}: U \to \mathbb{C}$ by \begin{align*} T^C(\partial_{z_i},\partial_{z_j})=\sum_{p=1}^n T^p_{ij}\partial_{z_p}. \end{align*} Since the coordinate vector fields commute, the torsion coefficients are \begin{align*} T^p_{ij}=\Gamma^p_{ij}-\Gamma^p_{ji}=\sum_{q=1}^n h^{p\bar q}\left(\partial_{z_i}h_{j\bar q}-\partial_{z_j}h_{i\bar q}\right). \end{align*} [/step] [step:Compute the $(2,1)$-part of $d\omega$] Since $\omega$ has type $(1,1)$, its [exterior derivative](/theorems/1525) decomposes by type as \begin{align*} d\omega=\partial\omega+\bar\partial\omega, \end{align*} where $\partial\omega$ has type $(2,1)$ and $\bar\partial\omega$ has type $(1,2)$. Using $\partial dz_j=0$ and $\partial d\bar z_k=0$, we get \begin{align*} \partial\omega=i\sum_{i,j,k=1}^n \partial_{z_i}h_{j\bar k}\, dz_i\wedge dz_j\wedge d\bar z_k. \end{align*} Terms with $i=j$ vanish because $dz_i\wedge dz_i=0$. Grouping the terms with indices $i<j$, and using $dz_j\wedge dz_i=-dz_i\wedge dz_j$, gives \begin{align*} \partial\omega=i\sum_{1\le i<j\le n}\sum_{k=1}^n \left(\partial_{z_i}h_{j\bar k}-\partial_{z_j}h_{i\bar k}\right)dz_i\wedge dz_j\wedge d\bar z_k. \end{align*} Therefore $\partial\omega=0$ on $U$ if and only if \begin{align*} \partial_{z_i}h_{j\bar k}-\partial_{z_j}h_{i\bar k}=0 \end{align*} for every $1\le i<j\le n$ and every $1\le k\le n$. [guided] The point of this calculation is to isolate exactly which first derivatives of the metric coefficients are measured by $d\omega$. Because $\omega$ is a $(1,1)$-form, the exterior derivative splits into its two possible type components: \begin{align*} d\omega=\partial\omega+\bar\partial\omega. \end{align*} The component $\partial\omega$ has type $(2,1)$, so it differentiates only the coefficient functions $h_{j\bar k}$ in holomorphic directions. Since \begin{align*} \omega=i\sum_{j,k=1}^n h_{j\bar k}\, dz_j\wedge d\bar z_k, \end{align*} and since the coordinate one-forms $dz_j$ and $d\bar z_k$ are constant in these coordinates, applying $\partial$ gives \begin{align*} \partial\omega=i\sum_{i,j,k=1}^n \partial_{z_i}h_{j\bar k}\, dz_i\wedge dz_j\wedge d\bar z_k. \end{align*} Now we collect the coefficient of each basis form $dz_i\wedge dz_j\wedge d\bar z_k$ with $i<j$. The summand with indices $(i,j,k)$ contributes \begin{align*} i\partial_{z_i}h_{j\bar k}\, dz_i\wedge dz_j\wedge d\bar z_k. \end{align*} The summand with indices $(j,i,k)$ contributes \begin{align*} i\partial_{z_j}h_{i\bar k}\, dz_j\wedge dz_i\wedge d\bar z_k=-i\partial_{z_j}h_{i\bar k}\, dz_i\wedge dz_j\wedge d\bar z_k. \end{align*} Adding these two contributions gives the coefficient \begin{align*} i\left(\partial_{z_i}h_{j\bar k}-\partial_{z_j}h_{i\bar k}\right). \end{align*} Thus \begin{align*} \partial\omega=i\sum_{1\le i<j\le n}\sum_{k=1}^n \left(\partial_{z_i}h_{j\bar k}-\partial_{z_j}h_{i\bar k}\right)dz_i\wedge dz_j\wedge d\bar z_k. \end{align*} The forms $dz_i\wedge dz_j\wedge d\bar z_k$ with $i<j$ form a local frame for $(2,1)$-forms, so $\partial\omega=0$ is equivalent to the vanishing of every coefficient: \begin{align*} \partial_{z_i}h_{j\bar k}-\partial_{z_j}h_{i\bar k}=0. \end{align*} [/guided] [/step] [step:Identify the coefficients of $\partial\omega$ with Chern torsion] For fixed indices $i$ and $j$, define the covector coefficients $A_{ij\bar q}: U \to \mathbb{C}$ by \begin{align*} A_{ij\bar q}=\partial_{z_i}h_{j\bar q}-\partial_{z_j}h_{i\bar q}. \end{align*} The torsion formula from the first step gives \begin{align*} T^p_{ij}=\sum_{q=1}^n h^{p\bar q}A_{ij\bar q}. \end{align*} Because $(h^{p\bar q})$ is the inverse Hermitian metric matrix, multiplication by $(h^{p\bar q})$ is an isomorphism from covectors with barred index to vectors with holomorphic index. Hence \begin{align*} T^p_{ij}=0\ \text{for all }p \end{align*} if and only if \begin{align*} A_{ij\bar q}=0\ \text{for all }q. \end{align*} By the coefficient computation for $\partial\omega$, this is equivalent to $\partial\omega=0$ on $U$. Since $U$ was arbitrary, $T^C=0$ on $X$ if and only if $\partial\omega=0$ on $X$. [/step] [step:Use reality of $\omega$ to pass from $\partial\omega$ to $d\omega$] The fundamental form $\omega$ of a Hermitian metric is real, so complex conjugation of forms gives \begin{align*} \overline{\partial\omega}=\bar\partial\omega. \end{align*} Therefore $\partial\omega=0$ if and only if $\bar\partial\omega=0$. Since \begin{align*} d\omega=\partial\omega+\bar\partial\omega \end{align*} and the summands have different bidegrees, $d\omega=0$ if and only if both $\partial\omega=0$ and $\bar\partial\omega=0$. Combining this with the previous step gives \begin{align*} T^C=0 \iff \partial\omega=0 \iff d\omega=0. \end{align*} This proves that the Chern torsion vanishes exactly when the associated Hermitian form is closed, which is the Kähler condition. [/step]