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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!

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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.

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  • $\begingroup$ Why must $S_1$ and $S_2$ be isomorphic if $U_1$ and $U_2$ are isomorphic? $\endgroup$
    – Yuan Yang
    Jul 4 at 21:04
  • $\begingroup$ Let $\tilde{U}_i:=\pi _i^{-1}(U_i) $. Since $\pi _1(U_i)=\mathbb{Z}/4$, the étale double covering $\tilde{U}_i\rightarrow U_i$ is canonical, so $\tilde{U}_1\cong \tilde{U}_2$. Then $S_1\cong S_2$ because $S_i$ is the normalization of $\mathbb{P}^2$ in $\tilde{U} _i$ (stacks project, 29.53). $\endgroup$
    – abx
    Jul 5 at 3:49
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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.

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  • $\begingroup$ This is incisive and elegant! I don't even know if Mr.Yoshihara noticed this when he wrote his papers. He dealt with some more non-smooth cases though. $\endgroup$
    – Yuan Yang
    Jul 4 at 20:44
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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.

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  • $\begingroup$ Indeed. If we allow these hypersurfaces to be reducible it seems their complements are less likely to be isomorphic. $\endgroup$
    – Yuan Yang
    Jul 4 at 20:49

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