[proofplan] The identity is local, because the Kodaira-Spencer bracket, contraction with $\Omega$, and the operator $\partial$ are all defined by tensorial or differential operations that agree on overlaps. We therefore work in one holomorphic coordinate chart and keep the local holomorphic coefficient of $\Omega$ throughout the computation. The proof is a direct coordinate expansion: first we write the two hypotheses $\partial(\varphi\lrcorner\Omega)=0$ and $\partial(\psi\lrcorner\Omega)=0$ as weighted divergence identities, then we compute $\partial(\varphi\lrcorner(\psi\lrcorner\Omega))$ and use those weighted identities to cancel the extra terms. The remaining terms are exactly the contraction of the Kodaira-Spencer bracket with $\Omega$ under the stated contraction and bracket conventions. [/proofplan] [step:Reduce the identity to a holomorphic coordinate computation] Fix a holomorphic coordinate chart $(U,z_1,\dots,z_n)$ on $X$. Since $\Omega$ is nowhere vanishing and holomorphic, there is a nowhere-zero [holomorphic function](/page/Holomorphic%20Function) \begin{align*} f:U\to \mathbb C \end{align*} such that \begin{align*} \Omega=f\,dz_1\wedge\cdots\wedge dz_n. \end{align*} We keep this coefficient in the calculation; for the holomorphic operator $\partial$, the term $\partial f$ contributes to the same weighted divergence expressions as the derivatives of the coefficients of $\varphi$ and $\psi$. Write \begin{align*} \varphi=\sum_{i=1}^n \varphi_i\otimes \partial_{z_i} \end{align*} and \begin{align*} \psi=\sum_{j=1}^n \psi_j\otimes \partial_{z_j}, \end{align*} where each \begin{align*} \varphi_i:U\to \Lambda^{0,1}T^*U \end{align*} and each \begin{align*} \psi_j:U\to \Lambda^{0,1}T^*U \end{align*} is a smooth $(0,1)$-form. Let \begin{align*} \Omega_0:=dz_1\wedge\cdots\wedge dz_n. \end{align*} For each $i$, define the $(n-1,0)$-form \begin{align*} \Omega_i:=\partial_{z_i}\lrcorner\Omega_0. \end{align*} For each pair $i,j$, define the $(n-2,0)$-form \begin{align*} \Omega_{ij}:=\partial_{z_i}\lrcorner(\partial_{z_j}\lrcorner\Omega_0). \end{align*} The contraction convention in the statement gives \begin{align*} \varphi\lrcorner\Omega=f\sum_{i=1}^n \varphi_i\wedge \Omega_i \end{align*} and the stated iterated-contraction convention gives \begin{align*} \varphi\lrcorner(\psi\lrcorner\Omega)=f\sum_{i,j=1}^n \varphi_i\wedge\psi_j\wedge\Omega_{ij}. \end{align*} For a smooth $(0,1)$-form $\alpha$ on $U$, the notation $\partial_{z_i}\alpha$ denotes the smooth $(0,1)$-form obtained by differentiating the coefficient functions of $\alpha$ in the $z_i$ direction. [/step] [step:Convert the closedness hypotheses into weighted local divergence identities] Since each $\Omega_i$ has constant holomorphic coefficients in the chosen chart, applying $\partial$ gives \begin{align*} \partial(\varphi\lrcorner\Omega)=\sum_{i=1}^n \partial(f\varphi_i)\wedge\Omega_i. \end{align*} Write $\partial(f\varphi_i)$ as the sum of its holomorphic coordinate components. When this $(1,1)$-form is wedged with $\Omega_i$, every component vanishes except the component containing $dz_i$, and the signs are exactly those built into the definitions $\Omega_i=\partial_{z_i}\lrcorner\Omega_0$. Hence the identity $\partial(\varphi\lrcorner\Omega)=0$ is equivalent to \begin{align*} \sum_{i=1}^n \partial_{z_i}(f\varphi_i)=0. \end{align*} Similarly, the hypothesis $\partial(\psi\lrcorner\Omega)=0$ gives \begin{align*} \sum_{i=1}^n \partial_{z_i}(f\psi_i)=0. \end{align*} [guided] The role of the assumptions is to remove the terms in the product rule where $\partial$ differentiates the coefficient multiplying the contracted coordinate direction. We must keep the holomorphic factor $f$ from $\Omega=f\Omega_0$, because $\partial(f\alpha)$ contains the term $(\partial f)\wedge\alpha$. For each $i$, the form $\Omega_i=\partial_{z_i}\lrcorner\Omega_0$ is the coordinate $(n-1,0)$-form obtained from $\Omega_0$ by omitting $dz_i$ with the sign fixed by the contraction convention. If $\alpha_i$ is any smooth $(0,1)$-form, then $\partial(f\alpha_i)$ is a smooth $(1,1)$-form. Its component containing $dz_k$ wedges to zero with $\Omega_i$ whenever $k\ne i$, because then $dz_k$ already occurs in $\Omega_i$. The only surviving component is therefore the $dz_i$ component, namely $dz_i\wedge\partial_{z_i}(f\alpha_i)$, and the contraction sign gives the common coordinate-volume factor. Applying this with $\alpha_i=\varphi_i$ yields \begin{align*} \partial(\varphi\lrcorner\Omega)=0 \end{align*} if and only if \begin{align*} \sum_{i=1}^n \partial_{z_i}(f\varphi_i)=0. \end{align*} The same coefficient calculation with $\alpha_i=\psi_i$ yields \begin{align*} \sum_{i=1}^n \partial_{z_i}(f\psi_i)=0. \end{align*} These are weighted divergences, not unweighted divergences; the derivatives of $f$ are part of the cancellation used in the next step. [/guided] [/step] [step:Expand the derivative of the double contraction and identify the bracket terms] Using the graded Leibniz rule for $\partial$ and the fact that each $\Omega_{ij}$ has constant holomorphic coefficients in the coordinate chart, we obtain \begin{align*} \partial\left(\varphi\lrcorner(\psi\lrcorner\Omega)\right)=\sum_{i,j=1}^n \partial(f\varphi_i\wedge\psi_j)\wedge\Omega_{ij}. \end{align*} Since $f\varphi_i$ has total degree $1$, the graded Leibniz rule gives \begin{align*} \partial(f\varphi_i\wedge\psi_j)=\partial(f\varphi_i)\wedge\psi_j-f\varphi_i\wedge\partial\psi_j. \end{align*} The contractions satisfy the following sign identities: $dz_i\wedge\Omega_i=\Omega_0$, $dz_i\wedge\Omega_{ij}=\Omega_j$, and $dz_j\wedge\Omega_{ij}=-\Omega_i$. These follow directly from $\Omega_i=\partial_{z_i}\lrcorner\Omega_0$ and $\Omega_{ij}=\partial_{z_i}\lrcorner(\partial_{z_j}\lrcorner\Omega_0)$; if $i=j$, then $\Omega_{ij}=0$, so the same displayed identities give zero contribution to the double sum. Expanding $\partial(f\varphi_i)$ and $\partial\psi_j$ in holomorphic coordinate components and using these three identities gives \begin{align*} \partial\left(\varphi\lrcorner(\psi\lrcorner\Omega)\right)=f\sum_{i,j=1}^n\left(\varphi_i\wedge\partial_{z_i}\psi_j+\psi_i\wedge\partial_{z_i}\varphi_j\right)\wedge\Omega_j+\sum_{j=1}^n\left(\sum_{i=1}^n\partial_{z_i}(f\varphi_i)\right)\wedge\psi_j\wedge\Omega_j-\sum_{i=1}^n\varphi_i\wedge\left(\sum_{j=1}^n\partial_{z_j}(f\psi_j)\right)\wedge\Omega_i. \end{align*} The two final sums vanish by the weighted divergence identities proved in the previous step. Hence \begin{align*} \partial\left(\varphi\lrcorner(\psi\lrcorner\Omega)\right)=f\sum_{i,j=1}^n\left(\varphi_i\wedge\partial_{z_i}\psi_j+\psi_i\wedge\partial_{z_i}\varphi_j\right)\wedge\Omega_j. \end{align*} By the stated local convention for the Kodaira-Spencer bracket, \begin{align*} [\varphi,\psi]=\sum_{i,j=1}^n\left(\varphi_i\wedge \partial_{z_i}\psi_j+\psi_i\wedge \partial_{z_i}\varphi_j\right)\otimes \partial_{z_j}. \end{align*} Contracting this expression with $\Omega=f\Omega_0$ gives \begin{align*} [\varphi,\psi]\lrcorner\Omega=f\sum_{i,j=1}^n\left(\varphi_i\wedge \partial_{z_i}\psi_j+\psi_i\wedge \partial_{z_i}\varphi_j\right)\wedge\Omega_j. \end{align*} Therefore \begin{align*} \partial\left(\varphi\lrcorner(\psi\lrcorner\Omega)\right)=[\varphi,\psi]\lrcorner\Omega. \end{align*} [guided] We now do the sign-sensitive part of the computation explicitly. The point is that $\Omega_{ij}$ already contains every holomorphic coordinate differential except $dz_i$ and $dz_j$, so after wedging with $\Omega_{ij}$ only the $dz_i$ and $dz_j$ components of a holomorphic differential can survive. The contraction definitions give three elementary sign rules: \begin{align*} dz_i\wedge\Omega_i=\Omega_0. \end{align*} Also, \begin{align*} dz_i\wedge\Omega_{ij}=\Omega_j \end{align*} and \begin{align*} dz_j\wedge\Omega_{ij}=-\Omega_i. \end{align*} These are obtained by writing $\Omega_i=\partial_{z_i}\lrcorner\Omega_0$ and $\Omega_{ij}=\partial_{z_i}\lrcorner(\partial_{z_j}\lrcorner\Omega_0)$ and moving the displayed $dz_i$ or $dz_j$ back to its coordinate position in $dz_1\wedge\cdots\wedge dz_n$. If $i=j$, then the double contraction is zero, so that summand contributes nothing. Using the graded Leibniz rule for $\partial$ on the product $f\varphi_i\wedge\psi_j$, where $f\varphi_i$ has total degree $1$, gives \begin{align*} \partial(f\varphi_i\wedge\psi_j)=\partial(f\varphi_i)\wedge\psi_j-f\varphi_i\wedge\partial\psi_j. \end{align*} Next expand the holomorphic differentials coefficientwise: \begin{align*} \partial(f\varphi_i)=\sum_{k=1}^n dz_k\wedge\partial_{z_k}(f\varphi_i) \end{align*} and \begin{align*} \partial\psi_j=\sum_{k=1}^n dz_k\wedge\partial_{z_k}\psi_j. \end{align*} When wedged with $\Omega_{ij}$, all $k\notin\{i,j\}$ terms vanish because $dz_k$ is already a factor of $\Omega_{ij}$. Applying the three sign rules above and then adding and subtracting the $\partial_{z_j}f$ terms so that the second divergence appears as $\partial_{z_j}(f\psi_j)$ yields \begin{align*} \partial\left(\varphi\lrcorner(\psi\lrcorner\Omega)\right)=f\sum_{i,j=1}^n\left(\varphi_i\wedge\partial_{z_i}\psi_j+\psi_i\wedge\partial_{z_i}\varphi_j\right)\wedge\Omega_j+\sum_{j=1}^n\left(\sum_{i=1}^n\partial_{z_i}(f\varphi_i)\right)\wedge\psi_j\wedge\Omega_j-\sum_{i=1}^n\varphi_i\wedge\left(\sum_{j=1}^n\partial_{z_j}(f\psi_j)\right)\wedge\Omega_i. \end{align*} The previous step proved exactly that \begin{align*} \sum_{i=1}^n\partial_{z_i}(f\varphi_i)=0 \end{align*} and \begin{align*} \sum_{j=1}^n\partial_{z_j}(f\psi_j)=0. \end{align*} Thus both divergence sums vanish, leaving \begin{align*} \partial\left(\varphi\lrcorner(\psi\lrcorner\Omega)\right)=f\sum_{i,j=1}^n\left(\varphi_i\wedge\partial_{z_i}\psi_j+\psi_i\wedge\partial_{z_i}\varphi_j\right)\wedge\Omega_j. \end{align*} By the bracket convention in the theorem statement, \begin{align*} [\varphi,\psi]=\sum_{i,j=1}^n\left(\varphi_i\wedge \partial_{z_i}\psi_j+\psi_i\wedge \partial_{z_i}\varphi_j\right)\otimes \partial_{z_j}. \end{align*} Contracting this with $\Omega=f\Omega_0$ converts each $\partial_{z_j}$ into $\Omega_j$ and gives \begin{align*} [\varphi,\psi]\lrcorner\Omega=f\sum_{i,j=1}^n\left(\varphi_i\wedge \partial_{z_i}\psi_j+\psi_i\wedge \partial_{z_i}\varphi_j\right)\wedge\Omega_j. \end{align*} Therefore \begin{align*} \partial\left(\varphi\lrcorner(\psi\lrcorner\Omega)\right)=[\varphi,\psi]\lrcorner\Omega. \end{align*} [/guided] [/step] [step:Pass from the local coordinate identity to the global equality] The previous step proves \begin{align*} [\varphi,\psi]\lrcorner\Omega=\partial\left(\varphi\lrcorner(\psi\lrcorner\Omega)\right) \end{align*} on the chosen coordinate chart $U$, with the local coefficient $f$ already included. The two sides are globally defined smooth $(n-1,2)$-forms on $X$. Since the equality holds in every holomorphic coordinate chart, the local identities agree on overlaps and therefore define a global equality on $X$. This proves the Tian-Todorov identity. [/step]