TITLE: A characterisation of $\mathbb{P}^n$
QUESTION [24 upvotes]: 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.

REPLY [9 votes]: Here's a quick proof over a finite field $\mathbb F_q$, if we assume that the isomorphisms with $\mathbb A^n$ are defined over $\mathbb F_q$. Suppose $X$ of dimension $n$ embedded in $\mathbb P^N_{\mathbb F_q} $ has this property. Every hyperplane complement in $\mathbb P^N(\mathbb F_q)$ contains $q^N$ points, of which $q^n$ lie in $X(\mathbb F_q)$. So the fraction of points in the hyperplane complement that also lie in $X(\mathbb F_q) $ is $q^{n-N}$. Averaging over all hyperplane complements, as each point in $\mathbb P^{N}(\mathbb F_q)$ is in the same number of hyperplane complements, we see that the fraction of points in $\mathbb P^N(\mathbb F_q)$ that also lie in $X(\mathbb F_q)$ is $q^{n-N}$. Thus $$|X(\mathbb F_q) |= q^{n-N} |\mathbb P^N(\mathbb F_q)| = q^n + q^{n-1} + \dots q^{n-N}$$ which is not an integer unless $N \leq n$, in which case $X = \mathbb P^n$ because it is an $n$-dimensional subscheme of $\mathbb P^N$.
If we replace the assumption by the statement that every hyperplane complement defined over $\mathbb F_q$ has an isomorphism with $\mathbb A^n$ defined over $\overline{\mathbb F}_q$, we can do it with a quick etale cohomology lemma.

Every variety $Y$ over $\mathbb F_q$ with $Y_{\overline{\mathbb F}_q} \cong \mathbb A^n_{\overline{\mathbb F}_q}$ has $Y(\mathbb F_q) = q^n$.

Proof: By the Grothendieck-Lefschetz fixed-point formula and the calculation of the cohomology of affine space we have $$Y(\mathbb F_q) =\sum_{i}(-1)^i \operatorname{tr}(\operatorname{Frob}_q, H^i_c(X, \mathbb Q_{\ell}) = \operatorname{tr} (\operatorname{Frob}_q, H^{2n}_c(\mathbb Q_{\ell})).$$ Furthermore, $H^{2n}_c$ is one-dimensional and the Frobenius eigenvalue on it can be computed by Poincare duality as $q^n$, because the unique Frobenius eigenvalue on $H^0(Y_{\overline{\mathbb F}_q}, \mathbb Q_\ell)$ must be one.
With this enhanced form, we can get the statement over an arbitrary field by a spreading out argument. First, define a stratification of the dual variety, and, over a finite etale covering each stratum, an explicit isomorphism from the family of hyperplane complements to $\mathbb A^n$. To do this, choose an isomorphism at the geometric generic point, which automatically extends to a finite etale cover of an open set, and induct. Next, choose a finitely generated ring over which $X$, the projective embedding, this stratification, the finite etale coverings, and all these isomorphisms are defined. Over an open subset, the stratification will remain a stratification and the isomorphisms will remain isomorphisms. Choosing a finite field-valued point in that subset and performing a counting argument, we see $n=N$ and conclude $X = \mathbb P^n= \mathbb P^N$.