3
$\begingroup$

Let $f: X\dashrightarrow\mathbb{CP}^n$ be a birational map where $X$ is a smooth, projective variety. Then there are some closed subvarieties $Z'\subset X$ and $Z \subset \mathbb{CP}^n$ such that $f$ induces an isomorphism $X \setminus Z' \cong \mathbb{CP}^n \setminus Z.$ Can we assume $Z \subset \mathbb{CP}^n$ has codimension at least two? My intuition is that no codimension one things in $\mathbb{CP}^n$ can be "contracted."

$\endgroup$
1
  • 1
    $\begingroup$ I changed $f:X-->\mathbb{CP}^n$ to $f:X\dashrightarrow\mathbb{CP}^n.$ I hope that is what was intended. $\endgroup$ Commented May 17 at 18:23

1 Answer 1

9
$\begingroup$

This not true. The simplest counterexample is the linear projection $Q \dashrightarrow \mathbb{P}^2$ from a smooth quadric surface $Q \subset \mathbb{P}^3$ from a point. The inverse map blows up two points on $\mathbb{P}^2$ and contracts the line connecting them. So, to obtain an isomorphism one has to remove this line.

$\endgroup$
1
  • $\begingroup$ Thank you, I should have considered the surface case. $\endgroup$ Commented May 17 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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