[proofplan] The infinitesimal period map formula identifies the first-order variation of the Hodge line with contraction of the Kodaira-Spencer class against the holomorphic volume form. It remains to prove that this contraction map has no kernel. Since $\Omega$ is nowhere vanishing, contraction with $\Omega$ identifies the holomorphic tangent bundle $T_X$ with $\Omega_X^{n-1}$, and because $\Omega$ is holomorphic this identification commutes with the Dolbeault operators. Passing to Dolbeault cohomology gives an isomorphism $H^1(X,T_X)\cong H^{n-1,1}(X)$, so the $H^{n-1,1}$ component of $d\mathcal P_0$ is injective. [/proofplan]

[step:Identify the period differential with contraction by the holomorphic volume form]Let \begin{align*} \rho_0:T_0S \longrightarrow H^1(X,T_X) \end{align*} denote the Kodaira-Spencer map of the deformation at $0\in S$. By hypothesis, $\rho_0$ is an isomorphism. Thus every tangent vector $v\in T_0S$ has a unique Kodaira-Spencer class \begin{align*} \rho_0(v)=[\varphi]\in H^1(X,T_X). \end{align*} The hypotheses of the infinitesimal period map formula [citetheorem:9141] are satisfied: the central fibre $X$ is compact Kähler, the family is marked so the period map is expressed in the fixed [vector space](/page/Vector%20Space) $H^n(X,\mathbb C)$, the class $\rho_0(v)$ is represented by a Dolbeault $T_X$-valued $(0,1)$-form $\varphi$, and $\Omega$ is a nowhere-vanishing holomorphic section of $K_X$. Therefore the $H^{n-1,1}(X)$ component of $d\mathcal P_0(v)$ is \begin{align*} [\varphi\lrcorner\Omega]\in H^{n-1,1}(X). \end{align*} Therefore, under the identification $T_0S\cong H^1(X,T_X)$ given by $\rho_0$, the relevant component of the period differential is the map \begin{align*} \kappa:H^1(X,T_X)&\longrightarrow H^{n-1,1}(X) \end{align*} defined by \begin{align*} \kappa([\varphi])&=[\varphi\lrcorner\Omega]. \end{align*}[/step]

[guided]The period map records the Hodge line $H^{n,0}$ inside the fixed marked cohomology space. Its derivative has several equivalent descriptions, but the part we need is its component in the first possible lower Hodge type, namely $H^{n-1,1}(X)$. Let \begin{align*} \rho_0:T_0S \longrightarrow H^1(X,T_X) \end{align*} be the Kodaira-Spencer map. The statement assumes that the deformation is locally complete in the needed sense: $\rho_0$ identifies the Zariski tangent space $T_0S$ with the full cohomology group $H^1(X,T_X)$. Now take a tangent vector $v\in T_0S$ and write \begin{align*} \rho_0(v)=[\varphi], \end{align*} where $\varphi$ is a Dolbeault representative in $A^{0,1}(X,T_X)$. We now check the hypotheses of the infinitesimal period map formula [citetheorem:9141]. The central fibre $X$ is compact Kähler by assumption. The family is marked, so the period map is written in the fixed cohomology vector space $H^n(X,\mathbb C)$. The Kodaira-Spencer class of $v$ is represented by the Dolbeault $T_X$-valued $(0,1)$-form $\varphi$, and $\Omega$ is a nowhere-vanishing holomorphic section of $K_X$. The formula therefore says precisely that the first variation of the holomorphic volume class has $H^{n-1,1}$ component \begin{align*} [\varphi\lrcorner\Omega]\in H^{n-1,1}(X). \end{align*} Thus, after replacing $T_0S$ by $H^1(X,T_X)$ through $\rho_0$, the map whose injectivity we must prove is \begin{align*} \kappa:H^1(X,T_X)&\longrightarrow H^{n-1,1}(X) \end{align*} with \begin{align*} \kappa([\varphi])&=[\varphi\lrcorner\Omega]. \end{align*} This reduces local Torelli to a purely cohomological statement: contraction with the nowhere-vanishing holomorphic $n$-form must be an isomorphism on first Dolbeault cohomology.[/guided]

[step:Construct the contraction isomorphism of holomorphic vector bundles]Define the holomorphic bundle map \begin{align*} C:T_X&\longrightarrow \Omega_X^{n-1} \end{align*} by \begin{align*} C(\xi)&=\xi\lrcorner\Omega \end{align*} for each local holomorphic vector field $\xi$ on $X$. We show that $C$ is an isomorphism. Fix a point $p\in X$. Since $\Omega_p$ is a nonzero alternating $n$-form on the complex vector space $T_pX$, the [linear map](/page/Linear%20Map) \begin{align*} C_p:T_pX&\longrightarrow \Lambda^{n-1}T_p^*X \end{align*} defined by \begin{align*} C_p(\xi_p)&=\xi_p\lrcorner\Omega_p \end{align*} is injective: if $\xi_p\ne 0$, choose vectors $e_2,\dots,e_n\in T_pX$ such that $\xi_p,e_2,\dots,e_n$ is a basis of $T_pX$, and then \begin{align*} (\xi_p\lrcorner\Omega_p)(e_2,\dots,e_n)=\Omega_p(\xi_p,e_2,\dots,e_n)\ne 0. \end{align*} The vector spaces $T_pX$ and $\Lambda^{n-1}T_p^*X$ both have dimension $n$, so $C_p$ is an isomorphism. Since this holds for every $p\in X$, the holomorphic bundle map $C:T_X\to\Omega_X^{n-1}$ is a holomorphic vector bundle isomorphism.[/step]

[guided]Contraction with a top-degree form should turn vector fields into $(n-1)$-forms, but we must check that this is an isomorphism at every point. Define the holomorphic bundle map \begin{align*} C:T_X&\longrightarrow \Omega_X^{n-1} \end{align*} by \begin{align*} C(\xi)&=\xi\lrcorner\Omega \end{align*} for each local holomorphic vector field $\xi$ on $X$. Fix $p\in X$. The fibre map is \begin{align*} C_p:T_pX&\longrightarrow \Lambda^{n-1}T_p^*X \end{align*} with \begin{align*} C_p(\xi_p)&=\xi_p\lrcorner\Omega_p. \end{align*} If $\xi_p\ne 0$, choose vectors $e_2,\dots,e_n\in T_pX$ such that $\xi_p,e_2,\dots,e_n$ is a basis of $T_pX$. Because $\Omega_p$ is a nonzero alternating $n$-form and the chosen vectors form a basis, we have \begin{align*} (\xi_p\lrcorner\Omega_p)(e_2,\dots,e_n)=\Omega_p(\xi_p,e_2,\dots,e_n)\ne 0. \end{align*} Thus $C_p(\xi_p)\ne 0$, so $C_p$ is injective. Both $T_pX$ and $\Lambda^{n-1}T_p^*X$ have complex dimension $n$, hence an injective linear map between them is an isomorphism. Since this holds at every point $p\in X$, $C:T_X\to\Omega_X^{n-1}$ is a holomorphic vector bundle isomorphism.[/guided]

[step:Extend contraction to an isomorphism of Dolbeault complexes]For each integer $q\ge 0$, let $A^{0,q}(X,T_X)$ denote the complex vector space of smooth $T_X$-valued $(0,q)$-forms on $X$, let $A^{n-1,q}(X)$ denote the complex vector space of smooth scalar-valued $(n-1,q)$-forms on $X$, and let $\bar\partial_{T_X}:A^{0,q}(X,T_X)\to A^{0,q+1}(X,T_X)$ denote the Dolbeault operator induced by the holomorphic structure on $T_X$. Define \begin{align*} C_q:A^{0,q}(X,T_X)&\longrightarrow A^{n-1,q}(X) \end{align*} by \begin{align*} C_q(\varphi)&=\varphi\lrcorner\Omega. \end{align*} Because $C:T_X\to\Omega_X^{n-1}$ is a holomorphic vector bundle isomorphism, each $C_q$ is an isomorphism of complex vector spaces. It remains to check compatibility with the Dolbeault operators. Let $(U,z)$ be a holomorphic coordinate chart with coordinates $z_1,\dots,z_n$, and write the holomorphic volume form on $U$ as \begin{align*} \Omega|_U=f\,dz_1\wedge\cdots\wedge dz_n, \end{align*} where $f:U\to\mathbb C$ is holomorphic and nowhere zero. A local element $\varphi\in A^{0,q}(U,T_X)$ can be written as a finite sum \begin{align*} \varphi=\sum_{i,J}\varphi_{i,J}\,d\bar z_J\otimes\partial_{z_i}, \end{align*} where $J$ ranges over increasing $q$-tuples of indices and each coefficient $\varphi_{i,J}:U\to\mathbb C$ is smooth. Since $\bar\partial f=0$, applying $\bar\partial$ to the coefficients gives \begin{align*} \bar\partial(\varphi\lrcorner\Omega)=(\bar\partial_{T_X}\varphi)\lrcorner\Omega. \end{align*} Thus the maps $C_q$ assemble into an isomorphism of Dolbeault complexes \begin{align*} (A^{0,*}(X,T_X),\bar\partial_{T_X})\cong (A^{n-1,*}(X),\bar\partial). \end{align*}[/step]

[guided]The purpose of this step is to justify that contraction is not merely a pointwise bundle isomorphism; it is compatible with the differential used to define Dolbeault cohomology. For every $q\ge 0$, let $A^{0,q}(X,T_X)$ be the complex vector space of smooth $T_X$-valued $(0,q)$-forms, let $A^{n-1,q}(X)$ be the complex vector space of smooth scalar-valued $(n-1,q)$-forms, and let $\bar\partial_{T_X}:A^{0,q}(X,T_X)\to A^{0,q+1}(X,T_X)$ be the Dolbeault operator induced by the holomorphic structure on $T_X$. Define \begin{align*} C_q:A^{0,q}(X,T_X)&\longrightarrow A^{n-1,q}(X) \end{align*} by \begin{align*} C_q(\varphi)&=\varphi\lrcorner\Omega. \end{align*} Since $C:T_X\to\Omega_X^{n-1}$ is already a holomorphic vector bundle isomorphism, tensoring with $(0,q)$-forms gives an isomorphism on smooth sections, so each $C_q$ is bijective. We must verify that this bijection sends $\bar\partial_{T_X}$-closed forms to $\bar\partial$-closed forms and $\bar\partial_{T_X}$-exact forms to $\bar\partial$-exact forms. This is equivalent to the identity \begin{align*} \bar\partial(\varphi\lrcorner\Omega)=(\bar\partial_{T_X}\varphi)\lrcorner\Omega \end{align*} for every $\varphi\in A^{0,q}(X,T_X)$. We verify the identity in local holomorphic coordinates. Let $(U,z)$ be a holomorphic coordinate chart with coordinates $z_1,\dots,z_n$. Since $\Omega$ is a holomorphic section of $K_X$, there is a [holomorphic function](/page/Holomorphic%20Function) $f:U\to\mathbb C$ such that \begin{align*} \Omega|_U=f\,dz_1\wedge\cdots\wedge dz_n. \end{align*} The nowhere-vanishing hypothesis says $f(p)\ne 0$ for every $p\in U$, but the commutation with $\bar\partial$ uses the holomorphicity condition $\bar\partial f=0$. Write a local $T_X$-valued $(0,q)$-form as \begin{align*} \varphi=\sum_{i,J}\varphi_{i,J}\,d\bar z_J\otimes\partial_{z_i}, \end{align*} where $J$ ranges over increasing $q$-tuples and each $\varphi_{i,J}:U\to\mathbb C$ is smooth. Contracting with $\Omega$ gives a finite sum whose coefficients are products of the smooth functions $\varphi_{i,J}$ with the holomorphic function $f$, multiplied by fixed coordinate forms. Applying $\bar\partial$ differentiates only in the anti-holomorphic directions. Since $\bar\partial f=0$ and the coordinate holomorphic forms are $\bar\partial$-closed, the only differentiated coefficients are the $\varphi_{i,J}$. Therefore \begin{align*} \bar\partial(\varphi\lrcorner\Omega)=(\bar\partial_{T_X}\varphi)\lrcorner\Omega. \end{align*} This proves that the family $(C_q)_{q\ge 0}$ is an isomorphism of complexes \begin{align*} (A^{0,*}(X,T_X),\bar\partial_{T_X})\cong (A^{n-1,*}(X),\bar\partial). \end{align*}[/guided]

[step:Pass to cohomology and prove injectivity]Taking cohomology in degree $1$ of the Dolbeault complex isomorphism constructed above gives an isomorphism \begin{align*} H^1(X,T_X)&\longrightarrow H^{n-1,1}(X) \end{align*} defined by \begin{align*} [\varphi]&\longmapsto [\varphi\lrcorner\Omega]. \end{align*} Equivalently, this is the Hodge-theoretic tangent-space isomorphism for Calabi-Yau deformations [citetheorem:9138]. Therefore the map $\kappa$ identified in the first step is an isomorphism, hence injective. If $v\in T_0S$ satisfies $d\mathcal P_0(v)=0$, then its $H^{n-1,1}(X)$ component is zero, so \begin{align*} \kappa(\rho_0(v))=0. \end{align*} Since $\kappa$ and $\rho_0$ are injective, this implies $v=0$. Hence $d\mathcal P_0$ is injective, completing the proof.[/step]

[guided]The complex isomorphism constructed in the previous step induces an isomorphism on cohomology in every degree. In degree $1$, this gives the map \begin{align*} H^1(X,T_X)&\longrightarrow H^{n-1,1}(X) \end{align*} defined by \begin{align*} [\varphi]&\longmapsto [\varphi\lrcorner\Omega]. \end{align*} Equivalently, this is the Hodge-theoretic tangent-space isomorphism for Calabi-Yau deformations [citetheorem:9138]. The first step identified the $H^{n-1,1}(X)$ component of $d\mathcal P_0$ with this cohomology map, denoted $\kappa$. Since the degree-$1$ cohomology map is an isomorphism, $\kappa$ is injective. Now suppose $v\in T_0S$ satisfies $d\mathcal P_0(v)=0$. Then every component of $d\mathcal P_0(v)$ is zero, so in particular its $H^{n-1,1}(X)$ component is zero. Under the Kodaira-Spencer identification this says \begin{align*} \kappa(\rho_0(v))=0. \end{align*} Because $\kappa$ is injective, $\rho_0(v)=0$. Because $\rho_0:T_0S\to H^1(X,T_X)$ is an isomorphism by hypothesis, it is injective, and therefore $v=0$. Thus the kernel of $d\mathcal P_0$ is zero, so $d\mathcal P_0$ is injective.[/guided]