[proofplan] Choose a nowhere-vanishing holomorphic volume form on $X$ and use contraction with it to identify $T_X$-valued forms with ordinary forms of complementary holomorphic degree. The Kuranishi construction reduces smoothness of the local deformation space to solving the Maurer-Cartan equation order by order with no obstruction term. The [Tian-Todorov identity](/theorems/9139) makes every bracket obstruction $\partial$-exact after contraction with the chosen holomorphic volume form, while the Kähler $\partial\bar\partial$-lemma turns simultaneous $\partial$-exactness and $\bar\partial$-closedness into $\bar\partial$-exactness. This removes all obstruction equations, and the standard elliptic majorant estimate in Kuranishi's construction upgrades the formal solution to a convergent analytic family. [/proofplan]

[step:Choose the differential graded Lie algebra controlling complex deformations] Let $A^{p,q}(X)$ denote the [Fréchet space](/page/Fr%C3%A9chet%20Space) of smooth ordinary $(p,q)$-forms on $X$, and let $A^{0,q}(X,T_X)$ denote the Fréchet space of smooth $(0,q)$-forms on $X$ with values in the holomorphic tangent bundle $T_X$. Define the Dolbeault operators as the maps $\partial: A^{p,q}(X)\to A^{p+1,q}(X)$ and $\bar\partial: A^{p,q}(X)\to A^{p,q+1}(X)$, and also $\bar\partial: A^{0,q}(X,T_X)\to A^{0,q+1}(X,T_X)$. We use the Kodaira-Spencer bracket as the map $[\cdot,\cdot]:A^{0,p}(X,T_X)\times A^{0,q}(X,T_X)\to A^{0,p+q}(X,T_X)$, defined as the standard Frölicher-Nijenhuis extension of the Lie bracket of local holomorphic vector fields to vector-valued forms, with wedge product in the form variables and the Koszul sign convention. In local holomorphic coordinates, if $\varphi=\sum_i\alpha_i\otimes\partial_{z_i}$ and $\psi=\sum_j\beta_j\otimes\partial_{z_j}$, then $[\varphi,\psi]$ is obtained by applying this graded derivation rule to the coefficient functions and the local frame fields; this is the differential graded [Lie algebra](/page/Lie%20Algebra) bracket governing deformations of complex structure. A small element $\varphi\in A^{0,1}(X,T_X)$ defines an integrable deformation of the complex structure precisely when it satisfies the Maurer-Cartan equation \begin{align*} \bar\partial\varphi=\frac{1}{2}[\varphi,\varphi]. \end{align*} The linearization of this equation at $\varphi=0$ is $\bar\partial\varphi=0$, and quotienting by infinitesimal coordinate changes identifies first-order deformations with $H^1(X,T_X)$, by the [[Kodaira-Spencer Correspondence](/theorems/9117)][citetheorem:9117]. The Kuranishi construction packages these equations into a local analytic space whose Zariski tangent space is $H^1(X,T_X)$, by the [Kuranishi Theorem][citetheorem:9119]. [/step]

[step:Use the holomorphic volume form to convert brackets into ordinary forms]Since $K_X\cong\mathcal O_X$, where $\mathcal O_X$ is the holomorphic structure sheaf of $X$, choose a nowhere-vanishing holomorphic volume form \begin{align*} \Omega\in H^0(X,K_X). \end{align*} For each $q$, define the contraction map $\iota_\Omega:A^{0,q}(X,T_X)\to A^{n-1,q}(X)$ by $\iota_\Omega(\varphi)=\varphi\lrcorner\Omega$. Because $\Omega$ is nowhere vanishing, $\iota_\Omega$ is an isomorphism of smooth vector bundles in each bidegree, hence an isomorphism on smooth sections. Since $\Omega$ is holomorphic, contraction commutes with $\bar\partial$: \begin{align*} (\bar\partial\varphi)\lrcorner\Omega=\bar\partial(\varphi\lrcorner\Omega). \end{align*} Thus a $T_X$-valued form is $\bar\partial$-exact if and only if its contraction with $\Omega$ is $\bar\partial$-exact.[/step]

[guided]The holomorphic volume form is the mechanism that turns the deformation problem into a problem about ordinary differential forms. Since $\Omega$ is nowhere vanishing, contraction with $\Omega$ gives, at each point $x\in X$, a pointwise linear isomorphism from $T_xX\otimes \Lambda^qT_x^{0,1,*}X$ to $\Lambda^{n-1}T_x^{1,0,*}X\otimes \Lambda^qT_x^{0,1,*}X$. Taking smooth sections gives the map $\iota_\Omega:A^{0,q}(X,T_X)\to A^{n-1,q}(X)$ defined by $\iota_\Omega(\varphi)=\varphi\lrcorner\Omega$. The condition that $\Omega$ is holomorphic means $\bar\partial\Omega=0$. Therefore, when $\bar\partial$ differentiates the contraction $\varphi\lrcorner\Omega$, the only contribution is from $\bar\partial\varphi$, so \begin{align*} \bar\partial(\varphi\lrcorner\Omega)=(\bar\partial\varphi)\lrcorner\Omega. \end{align*} This matters because obstruction terms naturally lie in $A^{0,2}(X,T_X)$, while the $\partial\bar\partial$-lemma applies to ordinary forms. The contraction map lets us move between these two settings without losing information.[/guided]

[step:Construct the formal Maurer-Cartan solution recursively] Let $m=\dim_{\mathbb C}H^1(X,T_X)$. Let $H^{n-1,1}(X)$ denote the Dolbeault cohomology group of ordinary $(n-1,1)$-forms. Choose the [harmonic representatives](/theorems/2747) of a basis of $H^{n-1,1}(X)$ for the Kähler metric on $X$, and use the contraction isomorphism $\iota_\Omega^{-1}$ to write them as \begin{align*} \eta_1,\dots,\eta_m\in A^{0,1}(X,T_X). \end{align*} Their cohomology classes form a basis of $H^1(X,T_X)$. Because the contracted forms $\eta_a\lrcorner\Omega$ are Dolbeault-harmonic ordinary $(n-1,1)$-forms on the compact Kähler manifold $X$, the [Kähler identities](/theorems/3853) imply that every such Dolbeault-harmonic form is also $\partial$-closed. Hence \begin{align*} \partial(\eta_a\lrcorner\Omega)=0 \end{align*} for every $1\le a\le m$. For $t=(t_1,\dots,t_m)\in\mathbb C^m$, define the degree-one polynomial map $\varphi_1:\mathbb C^m\to A^{0,1}(X,T_X)$ by \begin{align*} \varphi_1(t)=\sum_{a=1}^{m}t_a\eta_a. \end{align*} We seek a formal [power series](/page/Power%20Series) $\varphi(t)=\sum_{r=1}^{\infty}\varphi_r(t)$ with coefficients in $A^{0,1}(X,T_X)$, where each $\varphi_r:\mathbb C^m\to A^{0,1}(X,T_X)$ is a homogeneous polynomial map of degree $r$ in $t_1,\dots,t_m$, and where \begin{align*} \bar\partial\varphi_r(t)=\frac{1}{2}\sum_{a+b=r}[\varphi_a(t),\varphi_b(t)] \end{align*} for every $r\ge2$. Assume $\varphi_1(t),\dots,\varphi_{r-1}(t)$ have been chosen so that the Maurer-Cartan equation holds through degree $r-1$. Define the degree-$r$ obstruction polynomial map $B_r:\mathbb C^m\to A^{0,2}(X,T_X)$ by \begin{align*} B_r(t)=\frac{1}{2}\sum_{a+b=r}[\varphi_a(t),\varphi_b(t)]. \end{align*} The graded Jacobi identity and the lower-order Maurer-Cartan equations imply \begin{align*} \bar\partial B_r(t)=0. \end{align*} Thus the degree-$r$ equation is solvable precisely when $B_r(t)$ is $\bar\partial$-exact. [/step]

[step:Show each obstruction is $\bar\partial$-exact by the Tian-Todorov identity]If $n\le 1$, then $A^{0,2}(X,T_X)=0$, so every obstruction form $B_r(t)$ vanishes. In this case define $\varphi_r(t)=0$ for every $r\ge2$; then $\bar\partial\varphi_r(t)=B_r(t)$ and the recursion is complete. Hence assume $n\ge2$ for the remaining argument. The induction hypothesis includes \begin{align*} \partial(\varphi_a(t)\lrcorner\Omega)=0 \end{align*} for every $a<r$. For $a=1$ this was ensured by the harmonic choice above. For $a\ge2$, the previous recursive step produced an ordinary form $\gamma_a(t)\in A^{n-2,1}(X)$ satisfying $\varphi_a(t)\lrcorner\Omega=\partial\gamma_a(t)$, and hence $\partial(\varphi_a(t)\lrcorner\Omega)=\partial^2\gamma_a(t)=0$. Therefore, for every pair $a,b<r$, the hypotheses of the [Tian-Todorov Identity][citetheorem:9139] are satisfied, and the contracted bracket $[\varphi_a(t),\varphi_b(t)]\lrcorner\Omega$ is $\partial$-exact. Hence \begin{align*} B_r(t)\lrcorner\Omega \end{align*} is $\partial$-exact. Since $\bar\partial B_r(t)=0$ and contraction with $\Omega$ commutes with $\bar\partial$, the form $B_r(t)\lrcorner\Omega$ is also $\bar\partial$-closed. We now apply the compact Kähler $\partial\bar\partial$-lemma to the ordinary smooth form $B_r(t)\lrcorner\Omega\in A^{n-1,2}(X)$. The hypotheses are satisfied: $X$ is compact Kähler by assumption, the form is ordinary rather than $T_X$-valued after contraction with $\Omega$, it is $\partial$-exact by the Tian-Todorov identity, and it is $\bar\partial$-closed by the previous paragraph. Therefore the form is $\partial\bar\partial$-exact, so there exists an ordinary form $\gamma_r(t)\in A^{n-2,1}(X)$ such that \begin{align*} B_r(t)\lrcorner\Omega=\bar\partial\partial\gamma_r(t). \end{align*} Because $\iota_\Omega$ is an isomorphism, there exists a unique $\varphi_r(t)\in A^{0,1}(X,T_X)$ satisfying \begin{align*} \varphi_r(t)\lrcorner\Omega=\partial\gamma_r(t). \end{align*} Then \begin{align*} (\bar\partial\varphi_r(t))\lrcorner\Omega=\bar\partial(\varphi_r(t)\lrcorner\Omega)=\bar\partial\partial\gamma_r(t)=B_r(t)\lrcorner\Omega. \end{align*} Since contraction with $\Omega$ is injective, $\bar\partial\varphi_r(t)=B_r(t)$. Also $\partial(\varphi_r(t)\lrcorner\Omega)=\partial^2\gamma_r(t)=0$, so the induction hypothesis needed at the next order is preserved. This completes the inductive step.[/step]

[guided]At order $r$, the only possible obstruction is the form \begin{align*} B_r(t)=\frac{1}{2}\sum_{a+b=r}[\varphi_a(t),\varphi_b(t)]. \end{align*} The lower-order Maurer-Cartan equations and the graded Jacobi identity imply $\bar\partial B_r(t)=0$, so $B_r(t)$ determines a Dolbeault cohomology class in $H^2(X,T_X)$. If $n\le1$, then $A^{0,2}(X,T_X)=0$, so $B_r(t)=0$ and the obstruction vanishes. We therefore focus on $n\ge2$. The Calabi-Yau hypothesis kills the obstruction in higher dimension. The first-order terms were chosen so that $\eta_a\lrcorner\Omega$ are harmonic ordinary forms, hence $\partial(\eta_a\lrcorner\Omega)=0$ by the Kähler identities. For higher lower-order terms, the previous recursive step produced an ordinary form $\gamma_a(t)\in A^{n-2,1}(X)$ and defined $\varphi_a(t)$ by $\varphi_a(t)\lrcorner\Omega=\partial\gamma_a(t)$, so $\partial(\varphi_a(t)\lrcorner\Omega)=\partial^2\gamma_a(t)=0$. Thus the hypotheses of the [Tian-Todorov Identity][citetheorem:9139] are satisfied for each pair $\varphi_a(t),\varphi_b(t)$, and the identity gives that \begin{align*} [\varphi_a(t),\varphi_b(t)]\lrcorner\Omega \end{align*} is $\partial$-exact. Summing over all $a+b=r$, we find that $B_r(t)\lrcorner\Omega$ is $\partial$-exact. We also know $\bar\partial B_r(t)=0$. Since $\Omega$ is holomorphic, contraction commutes with $\bar\partial$, so \begin{align*} \bar\partial(B_r(t)\lrcorner\Omega)=(\bar\partial B_r(t))\lrcorner\Omega=0. \end{align*} Thus $B_r(t)\lrcorner\Omega$ is both $\partial$-exact and $\bar\partial$-closed. On a compact Kähler manifold, the $\partial\bar\partial$-lemma applies to ordinary smooth forms that are simultaneously $\partial$-exact and $\bar\partial$-closed. Those hypotheses have just been verified for $B_r(t)\lrcorner\Omega$, so the lemma gives a form $\gamma_r(t)\in A^{n-2,1}(X)$ with \begin{align*} B_r(t)\lrcorner\Omega=\bar\partial\partial\gamma_r(t). \end{align*} Now define $\varphi_r(t)$ by requiring \begin{align*} \varphi_r(t)\lrcorner\Omega=\partial\gamma_r(t). \end{align*} This uniquely defines $\varphi_r(t)$ because contraction with the nowhere-vanishing $\Omega$ is an isomorphism. Applying $\bar\partial$ gives \begin{align*} (\bar\partial\varphi_r(t))\lrcorner\Omega=\bar\partial(\varphi_r(t)\lrcorner\Omega)=\bar\partial\partial\gamma_r(t)=B_r(t)\lrcorner\Omega. \end{align*} Injectivity of contraction with $\Omega$ then yields $\bar\partial\varphi_r(t)=B_r(t)$, exactly the degree-$r$ Maurer-Cartan equation.[/guided]

[step:Conclude convergence and smoothness of the Kuranishi space]The preceding induction proves more than the existence of one formal choice of corrections: at every order $r$, the obstruction cocycle $B_r(t)\in A^{0,2}(X,T_X)$ has zero class in $H^2(X,T_X)$, because it is $\bar\partial$-exact. This statement is independent of the particular representative used to solve the lower-order Maurer-Cartan equations: changing lower-order lifts changes $B_r(t)$ by a $\bar\partial$-exact term, since it is the standard obstruction class for extending a solution from order $r-1$ to order $r$. Hence the order-$r$ homogeneous component of the Kuranishi obstruction map is zero. In the analytic Kuranishi construction, the [Kuranishi Theorem][citetheorem:9119] gives a germ cut out inside a neighbourhood of $0\in H^1(X,T_X)$ by an analytic obstruction map with values in $H^2(X,T_X)$. We use the standard compatibility built into the theorem: the degree-$r$ Taylor coefficient of this obstruction map is the intrinsic obstruction class for extending a Maurer-Cartan solution from order $r-1$ to order $r$, and the analytic construction converges after the usual Kuranishi gauge and Green-operator choice. The previous paragraph shows that this intrinsic class is zero at every order, independently of the chosen lower-order lift. Hence all homogeneous Taylor coefficients of the analytic obstruction map vanish. Since an analytic map whose Taylor series at the origin is identically zero vanishes on a sufficiently small neighbourhood of the origin, the Kuranishi obstruction map is identically zero near $0$. Therefore the Kuranishi germ is an open neighbourhood of $0$ in the [vector space](/page/Vector%20Space) $H^1(X,T_X)$, after possibly shrinking the representative of the germ. It is smooth at the point corresponding to $X$, and its Zariski tangent space is $H^1(X,T_X)$. Consequently every first-order deformation class in $H^1(X,T_X)$ extends to an actual analytic deformation of the complex structure of $X$.[/step]

[guided]The remaining issue is to connect the formal obstruction calculation with the analytic Kuranishi space. At order $r$, the expression \begin{align*} B_r(t)=\frac{1}{2}\sum_{a+b=r}[\varphi_a(t),\varphi_b(t)] \end{align*} is a $\bar\partial$-closed $T_X$-valued $(0,2)$-form. Its Dolbeault cohomology class in $H^2(X,T_X)$ is the obstruction to solving the degree-$r$ Maurer-Cartan equation. The previous step showed that $B_r(t)$ is actually $\bar\partial$-exact, so this obstruction class is zero. Why is this enough for the Kuranishi obstruction map, whose corrections are usually chosen by a Green operator rather than by the primitive $\gamma_r(t)$ above? The obstruction at degree $r$ is a cohomology class, not a particular representative. If two lower-order choices solve the Maurer-Cartan equation through degree $r-1$, the difference between their degree-$r$ obstruction representatives is $\bar\partial$-exact; hence both choices define the same class in $H^2(X,T_X)$. Thus the Tian-Todorov argument kills the intrinsic obstruction class, and it also kills the corresponding Taylor coefficient of the analytic Kuranishi obstruction map. The [Kuranishi Theorem][citetheorem:9119] supplies the analytic part of the argument: the local deformation space is the zero set of an analytic obstruction map defined near $0\in H^1(X,T_X)$ with values in $H^2(X,T_X)$. Its construction includes the Green-operator recursion and the elliptic estimates needed for convergence, and its degree-$r$ Taylor coefficient is the intrinsic obstruction class for extending a Maurer-Cartan solution from order $r-1$ to order $r$; this is why it is enough to compute the cohomology class of $B_r(t)$ rather than to use the particular Green-operator representatives appearing in one Kuranishi gauge. Since every such intrinsic class vanishes by the induction, every homogeneous Taylor coefficient of the analytic obstruction map is zero. An analytic map with zero Taylor series at the origin vanishes after shrinking to a sufficiently small neighbourhood of the origin. Therefore the Kuranishi base is locally the whole neighbourhood in $H^1(X,T_X)$, so it is smooth with tangent space $H^1(X,T_X)$, and every first-order deformation class integrates to an analytic deformation.[/guided]