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Let $\pi:X\to S$ be a smooth family of complex manifolds (in the sense of deformation theory) such that $\pi^{-1}(0)\cong\mathbb{C}^n$ and $S\subset \mathbb{C}$ is a neighborhood of $0$. Is $\pi$ trivial? That is, is $X\cong \mathbb{C}^n\times S$ possibly after shrinking $S$?

I know that a smooth family of compact complex manifolds with $H^1(M,\mathcal{T}_M)=0$ is trivial (where $M=\pi^{-1}(0)$) but I'm not sure whether this extends to this non-compact situation.

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Example. $\pi: \{(z,w)\in {\mathbb C}^2: |zw|<1\}\to {\mathbb C}$, $\pi(z,w)=z$.

Edit. Similarly, to get a nontrivial deformation of ${\mathbb C}^n$, consider $$ X=\{(z_0, z_1,...,z_n)\in {\mathbb C}^{n+1}: |z_0 z_1|<1\} $$ and let $\pi$ be the projection of $X$ to ${\mathbb C}$ which the 1-st coordinate line in ${\mathbb C}^{n+1}$. Then $\pi^{-1}(t)$ (for $t\ne 0$) will be biholomorphic to $B \times {\mathbb C}^{n-1}$, where $B\subset {\mathbb C}$ is an open disk. Hence, these fibers are not biholomorphic to $\pi^{-1}(0)={\mathbb C}^{n}$.

A better question, I think, is if every Stein manifold of positive dimension admits a nontrivial deformation. I suspect that this is known.

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