T¹-lifting principle for analytic deformations

By [F2], represent $X_n$ by $\xi (t)={\sum }_{j=1}^n{\xi }_j{t}^j\in L_X^1\otimes (t)A_n$. Its curvature vanishes modulo $t^{n+1}$; write $R(t)=d\xi +[\xi ,\xi ]\mathrm{/}2=t^{n+1}{R}_{n+1}$ modulo $t^{n+2}$. The relative class ${\rho }_n$ is represented in the tangent resolution on $X_{n-1}$ by ${\xi }^{\mathrm{\prime }}(t)$ modulo $t^n$, up to the fixed sign from dualizing the cotangent extension. To see this, choose the relative coordinates of [F2] and write overlaps $w_i=F_{ij}(w_j,t)$. The difference of local lifts of the relative differentials to absolute differentials has $dt$ coefficient ${\mathrm{\partial }}_t{F}_{ij}$; this is the defining extension cocycle. In a smooth trivialization its Dolbeault representative is the commutator of ${\mathrm{\partial }}_t$ with $\bar{\mathrm{\partial }}+\xi \mathrm{\partial }$, whose coefficient is ${\xi }^{\mathrm{\prime }}$. Dualizing or reversing the Čech differential changes both the class and a putative lift by the same sign. Thus the lifting question is exactly that for $[{\xi }^{\mathrm{\prime }}]$. All coefficients modulo $t^n$ are retained, rather than replacing this relative class by its central tangent vector.

If $X_{n+1}$ exists, make its smooth trivialization compatible with the chosen one on $X_n$: any mismatch is a gauge map from [F2], and its finite vector-field coefficients lift to $A_{n+1}$. Its representative then has the form $\tilde{\xi }=\xi +t^{n+1}a$. Differentiating its Maurer–Cartan equation gives $d_{\xi }({\xi }^{\mathrm{\prime }}+(n+1)t^{n}a)=0$ modulo $t^{n+1}$. This closed tangent-valued form represents a class over $X_n$ restricting to $[{\xi }^{\mathrm{\prime }}]$ over $X_{n-1}$. Reinsert the fixed cotangent sign from step1.1 to obtain the required lift of ${\rho }_n$.

Conversely let ${\rho }_n$ have a lift, and represent the corresponding lift of $[{\xi }^{\mathrm{\prime }}]$ by a $d_{\xi }$-closed $\eta \in L_X^1\otimes A_n$ using [F2]. Its reduction differs from ${\xi }^{\mathrm{\prime }}$ by $d_{\xi }b$ modulo $t^n$ for some $b\in L_X^0\otimes A_{n-1}$. Lift the finitely many coefficients of $b$ to $A_n$ and subtract $d_{\xi }b$ from $\eta$. Since $d_{\xi }^2=0$ modulo $t^{n+1}$, the adjusted representative remains closed and has the exact form $\eta ={\xi }^{\mathrm{\prime }}+t^{n}c$ for a smooth $c\in L_X^1$. Differentiating the curvature polynomial and using the symmetry of the bracket on degree-one elements gives $d_{\xi }{\xi }^{\mathrm{\prime }}=R^{\mathrm{\prime }}(t)=(n+1)t^n{R}_{n+1}$ modulo $t^{n+1}$. Also $d_{\xi }(t^{n}c)=t^{n}dc$ at this order because $\xi$ has no constant term in $t$. Closedness therefore implies $dc=-(n+1)R_{n+1}$.

Set $\tilde{\xi }=\xi +t^{n+1}c\mathrm{/}(n+1)$. Its curvature modulo $t^{n+2}$ is $t^{n+1}(R_{n+1}+dc\mathrm{/}(n+1))=0$ by step 2.2, so it is a Maurer–Cartan lift over $A_{n+1}$. The reverse equivalence in [F2] constructs an actual smooth proper analytic Artin deformation, and its restriction compatibility recovers the given $X_n$ with its marking. This proves sufficiency; step2.1 proves necessity. The argument includes $n=1$ with $A_0=\mathbb{C}$ and divides only by the nonzero integer $n+1$. No Kähler, projectivity or canonical-bundle hypothesis was used.