Let $X$ be a projective variety (so, with some embedding) with the following curious property: for every hyperplane section $H$, we have that $X-H \cong \mathbb{A}^n$. Then is $X$ necessarily isomorphic to $\mathbb{P}^n$?

Assuming we are over the complex numbers for simplicity, one can show that $X$ has the same Hodge diamond as $\mathbb{CP}^n$ using purely topological arguments (such as long exact sequence of a pair, duality as in the proof of Lefschetz hyperplane, etc.) but this does not rule out these potential fake projective spaces.

This question is motivated by the fact that spheres are characterized by a similar property; i.e., a closed oriented manifold that is contractible upon removal of any point will be homeomorphic to a sphere.

Let $X$ be a projective variety (so, with some (edit: fixed nondegenerate closed) embedding) with the following curious property: for every hyperplane section $H$, we have that $X-H \cong \mathbb{A}^n$. Then is $X$ necessarily isomorphic to $\mathbb{P}^n$?

Assuming we are over the complex numbers for simplicity, one can show that $X$ has the same Hodge diamond as $\mathbb{CP}^n$ using purely topological arguments (such as long exact sequence of a pair, duality as in the proof of Lefschetz hyperplane, etc.) but this does not rule out these potential fake projective spaces.

This question is motivated by the fact that spheres are characterized by a similar property; i.e., a closed oriented manifold that is contractible upon removal of any point will be homeomorphic to a sphere.