Open complement of hypersurfaces Let $k$ be an algebraically closed field. Let $H_1, H_2$ be two smooth hypersurfaces of the same degree $d$ in $P^n_k$. Let $U_1,U_2$ be their complements respectively. Are $U_1,U_2$ isomorphic as algebraic varieties?
In $n=1,d=1$ case this is true, because the complement of any point is isomorphic to $A^1$.
But $n=2$ case I guess this might be false. I want to prove that if $U_1,U_2$ are isomorphic then they must be induced by an automorphism of $P^n$, but this seems hard. I read something about the 'complement problem' on enter link description here, but this seems to be a more complicated question, and it focuses on $A^n$ instead.
Maybe the $n=2,d=3$ case is easier? In this case, elliptic curves are isomorphic if and only if they have the same $j$-invariant. Can we read this from its complement?
Are there any solutions/counterexamples? Any comments are welcome!
 A: The answer is no. Perhaps the simplest case is $n=2$, $d=4$. There is a unique double covering $\pi _i:S_i\rightarrow \mathbb{P}^2$ branched along $H_i$. If   $U_1$ and $U_2$ are isomorphic,  $S_1$ and $S_2$ are isomorphic; then $H_1$ and $H_2$ are isomorphic, because $H_i$ is the branch locus of the morphism $\pi _i$, which is given by the anticanonical system.
A: The easiest case is $n = 1$, $d = 4$. Indeed, the embeddings $U_i \to \mathbb{P}^1$ are canonical, hence an isomorphism $U_1 \cong U_2$ extends to an isomorphism of the ambient projective lines and induces an isomorphism
$$
\mathbb{P}^1 \setminus U_1 \cong \mathbb{P}^1 \setminus U_2.
$$
So, if the cross-ratio of the four points $\mathbb{P}^1 \setminus U_1$ differs from the cross-ration of $\mathbb{P}^1 \setminus U_2$,
there can't be such an isomorphism.
A: If $U_1$ and $U_2$ are isomorphic then $H_1$ and $H_2$ are equal in the Grothendieck ring of varieties and thus, by the Larsen-Lunts theorem, stably birational, which if $d>n$ implies that they are isomorphic.
This is probably extreme overkill, but it seems to handle some different cases than the other answers.
