[proofplan] We first check that contraction with the fixed holomorphic volume form sends a Kodaira-Spencer representative to a well-defined Dolbeault class in $H^{n-1,1}(X)$. Then we invoke the general infinitesimal period-map formula, which identifies the first-order variation of the Hodge filtration with contraction by the Kodaira-Spencer class. Specializing that formula to the summand $H^{n,0}(X)$ gives exactly the displayed map from $H^1(X,T_X)$ to $\operatorname{Hom}(H^{n,0}(X),H^{n-1,1}(X))$. [/proofplan]

[step:Show that contraction with $\Omega$ gives a well-defined class in $H^{n-1,1}(X)$]Let $\varphi\in A^{0,1}(X,T_X)$ be a smooth $T_X$-valued $(0,1)$-form satisfying $\bar\partial\varphi=0$. Define the contraction map \begin{align*} C_\Omega:A^{0,1}(X,T_X)&\to A^{n-1,1}(X) \end{align*} by \begin{align*} C_\Omega(\psi):=\psi\lrcorner\Omega \end{align*} for every $\psi\in A^{0,1}(X,T_X)$. Thus $C_\Omega(\varphi)=\varphi\lrcorner\Omega\in A^{n-1,1}(X)$. Since $\Omega$ is holomorphic, $\bar\partial\Omega=0$. The Dolbeault operator on contractions satisfies the graded Leibniz rule \begin{align*} \bar\partial(\varphi\lrcorner\Omega)=(\bar\partial\varphi)\lrcorner\Omega-\varphi\lrcorner(\bar\partial\Omega). \end{align*} Both terms on the right vanish, so $\bar\partial(\varphi\lrcorner\Omega)=0$. Thus $\varphi\lrcorner\Omega$ defines a Dolbeault cohomology class \begin{align*} [\varphi\lrcorner\Omega]\in H^1(X,\Omega_X^{n-1})\cong H^{n-1,1}(X). \end{align*} If $\varphi'$ is another representative of the same class in $H^1(X,T_X)$, then there exists a smooth section $\xi\in A^{0,0}(X,T_X)$ such that \begin{align*} \varphi'=\varphi+\bar\partial\xi. \end{align*} Applying the same graded Leibniz rule to $\xi\lrcorner\Omega$ gives \begin{align*} (\bar\partial\xi)\lrcorner\Omega=\bar\partial(\xi\lrcorner\Omega), \end{align*} because $\bar\partial\Omega=0$. Therefore \begin{align*} \varphi'\lrcorner\Omega-\varphi\lrcorner\Omega=\bar\partial(\xi\lrcorner\Omega), \end{align*} so the Dolbeault class $[\varphi\lrcorner\Omega]$ depends only on $[\varphi]\in H^1(X,T_X)$.[/step]

[guided]The first point is that the displayed formula has to make sense in cohomology, not just on representatives. Let $\varphi\in A^{0,1}(X,T_X)$ be a smooth $T_X$-valued $(0,1)$-form with $\bar\partial\varphi=0$. Contracting the tangent-vector component into the holomorphic $n$-form $\Omega$ defines a map \begin{align*} C_\Omega:A^{0,1}(X,T_X)&\to A^{n-1,1}(X) \end{align*} by \begin{align*} C_\Omega(\psi):=\psi\lrcorner\Omega \end{align*} for every $\psi\in A^{0,1}(X,T_X)$. In particular, $C_\Omega(\varphi)=\varphi\lrcorner\Omega$ is an ordinary $(n-1,1)$-form. We must check that this form is $\bar\partial$-closed. The compatibility of $\bar\partial$ with contraction gives the graded identity \begin{align*} \bar\partial(\varphi\lrcorner\Omega)=(\bar\partial\varphi)\lrcorner\Omega-\varphi\lrcorner(\bar\partial\Omega). \end{align*} The first term vanishes because $\varphi$ represents a Dolbeault cohomology class, hence $\bar\partial\varphi=0$. The second term vanishes because $\Omega$ is holomorphic, hence $\bar\partial\Omega=0$. Therefore \begin{align*} \bar\partial(\varphi\lrcorner\Omega)=0. \end{align*} Thus $\varphi\lrcorner\Omega$ determines a class in $H^1(X,\Omega_X^{n-1})$, which is the same Hodge summand as $H^{n-1,1}(X)$ for the compact Kähler manifold $X$. Now suppose we change the representative of the Kodaira-Spencer class. If $\varphi'$ and $\varphi$ represent the same class in $H^1(X,T_X)$, then their difference is $\bar\partial$-exact, so there is a smooth section $\xi\in A^{0,0}(X,T_X)$ satisfying \begin{align*} \varphi'=\varphi+\bar\partial\xi. \end{align*} Contracting with $\Omega$ gives \begin{align*} \varphi'\lrcorner\Omega-\varphi\lrcorner\Omega=(\bar\partial\xi)\lrcorner\Omega. \end{align*} Because $\bar\partial\Omega=0$, the graded Leibniz rule applied to $\xi\lrcorner\Omega$ gives \begin{align*} (\bar\partial\xi)\lrcorner\Omega=\bar\partial(\xi\lrcorner\Omega). \end{align*} So the difference between the two contracted forms is $\bar\partial$-exact. Hence the cohomology class $[\varphi\lrcorner\Omega]\in H^{n-1,1}(X)$ is independent of the chosen representative $\varphi$.[/guided]

[step:Apply the general infinitesimal period-map formula to the top Hodge summand] Let $v\in T_{[X]}\mathcal M$ be a tangent vector, and let \begin{align*} \operatorname{KS}_{[X]}:T_{[X]}\mathcal M\to H^1(X,T_X) \end{align*} denote the Kodaira-Spencer map of the family $\pi:\mathcal X\to\mathcal M$ at the fibre $X=\pi^{-1}([X])$. By the Kodaira-Spencer identification, write \begin{align*} \operatorname{KS}_{[X]}(v)=[\varphi] \end{align*} for a $\bar\partial$-closed representative $\varphi\in A^{0,1}(X,T_X)$. By the [[Kodaira-Spencer Stability Theorem for Kähler Manifolds](/theorems/9126)][citetheorem:9126], after shrinking $\mathcal M$ around $[X]$, every fibre of the smooth proper holomorphic submersion $\pi:\mathcal X\to\mathcal M$ is a compact Kähler manifold. The [[Formula for the Infinitesimal Period Map](/theorems/9131)][citetheorem:9131] therefore applies to this family with the chosen local marking of the flat local system $R^n\pi_*\mathbb C$. We use the sign convention for the Kodaira-Spencer class and contraction under which that formula reads $[\varphi]\mapsto(\alpha\mapsto[\varphi\lrcorner\alpha])$. Specializing the theorem to weight $n$ and to the Hodge summand $H^{n,0}(X)$ gives \begin{align*} d\mathcal P_{[X]}(v)(\alpha)=[\varphi\lrcorner\alpha] \end{align*} for every element $\alpha\in H^{n,0}(X)$, viewed as the Dolbeault class of a holomorphic $n$-form $\alpha\in H^0(X,K_X)$. Taking $\alpha=\Omega$ yields \begin{align*} d\mathcal P_{[X]}(v)(\Omega)=[\varphi\lrcorner\Omega]\in H^{n-1,1}(X). \end{align*} Since $v$ corresponds to $[\varphi]$ under the Kodaira-Spencer identification, this is precisely \begin{align*} d\mathcal P_{[X]}([\varphi])(\Omega)=[\varphi\lrcorner\Omega]. \end{align*} [/step]

[step:Identify the target with $\operatorname{Hom}(H^{n,0}(X),H^{n-1,1}(X))$] By [Griffiths transversality](/theorems/9129), recorded in [Griffiths Transversality][citetheorem:9129], the derivative of the period map sends the top Hodge filtration piece $F^nH^n(X,\mathbb C)=H^{n,0}(X)$ into the quotient \begin{align*} F^{n-1}H^n(X,\mathbb C)/F^nH^n(X,\mathbb C). \end{align*} For a compact Kähler manifold, the [Hodge decomposition](/theorems/2745) identifies this quotient with $H^{n-1,1}(X)$. Therefore the derivative restricts to a [linear map](/page/Linear%20Map) \begin{align*} d\mathcal P_{[X]}:H^1(X,T_X)\to \operatorname{Hom}\bigl(H^{n,0}(X),H^{n-1,1}(X)\bigr). \end{align*} The form $\Omega$ generates the one-dimensional space $H^{n,0}(X)$: if $\alpha\in H^0(X,K_X)$, then the nowhere-vanishing property of $\Omega$ gives a [holomorphic function](/page/Holomorphic%20Function) $f:X\to\mathbb C$ with $\alpha=f\Omega$, and compact connectedness of $X$ forces $f$ to be constant. The preceding step computes the derivative on this generator, and the first step shows that the resulting class depends only on $[\varphi]$. Hence the derivative of the period map is exactly the claimed map. [/step]