# 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!

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.

• Why must $S_1$ and $S_2$ be isomorphic if $U_1$ and $U_2$ are isomorphic? Jul 4 '21 at 21:04
• 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).
– abx
Jul 5 '21 at 3:49

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.

• 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. Jul 4 '21 at 20:44

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.

• Indeed. If we allow these hypersurfaces to be reducible it seems their complements are less likely to be isomorphic. Jul 4 '21 at 20:49