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](http://archive.numdam.org/item/SB_1994-1995__37__295_0/), 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!



Open complement of hypersurfaces
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.


