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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 trivial Kodaira-Spencer class$H^1(M,\mathcal{T}_M)=0$ is trivial, (where $M=\pi^{-1}(0)$) but I'm not sure whether this extends to this non-compact situation.

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 trivial Kodaira-Spencer class is trivial, but I'm not sure whether this extends to this non-compact situation.

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

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 trivial Kodaira-Spencer class is trivial, but I'm not sure whether this extends to this non-compact situation.