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.



Is $\mathbb{C}^n$ rigid?
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. 


