MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a smooth projective variety (over an algebraically closed field; it could be the field of complex numbers); $Z$ is its hyperplane section. When there exists an etale $U/X$ such that:

  1. $Z$ can be lifted to $U$.

  2. There exists a morphism from $U$ to a curve $C$ such that $Z$ is the preimage of some point of $C$.

share|cite|improve this question
I am not sure I understand the question. If you had such an etale neighborhood of $Z$, then the normal bundle of $Z$ in $U$ will be trivial. But the normal bundle to $Z$ in $U$ is isomorphic to the normal bundle of $Z$ in $X$, and $N_{Z/X}$ can never be trivial for a hyperplane section. – Tony Pantev Sep 22 '12 at 14:51
up vote 3 down vote accepted

If $X$ is a curve, then just take linear projection from a disjoint codimension 2 linear subspace in the spanning hyperplane of $Z$. If the dimension of $X$ is $2$ or more, then there never exists such an étale neighborhood and morphism. For $Z$ of dimension $n-1 \geq 1$, for the normal sheaf $\mathcal{N}_{Z/X}$, the top intersection number $c_1(\mathcal{N}_{Z/X} )^{n-1}$ equals the degree of $X$, which is positive. Since $\mathcal{N}_{Z/X}$ equals $\mathcal{N}_{Z/U}$, also $c_1(\mathcal{N}_{Z/U})^{n-1}$ is positive. However, for a fiber of a morphism, $\mathcal{N}_{Z/U}$ is isomorphic to $\mathcal{O}_Z$ so that $c_1(\mathcal{N}_{Z/U})$ equals $0$.

share|cite|improve this answer

Suppose that $H$ is an hyperplane such that $Z = H \cap X$. Let $L$ be a subspace of codimension 1 of $H$; the pencil of hyperplanes passing through $L$ defines a morphism $X \smallsetminus L \to \mathbb P^1$ such that the inverse image of a point is $Z \smallsetminus L$. If you want $U$ to surject onto $X$ you can cover $X$ with various $X \smallsetminus L$, and take the disjoint union.

[Edit]: I misunderstood, the OP wants $Z$ to be embedded in $U$. This implies that the normal bundle of $Z$ in $X$ is trivial, and this never happens, unless $X$ is a curve.

share|cite|improve this answer
No, I don't need a surjection of $U$ onto $X$. On the other hand, I need a closed embedding $Z\to U$; $Z\setminus L\subset U$ is not enough. – Mikhail Bondarko Sep 22 '12 at 14:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.