1
$\begingroup$

Let $X$ be a smooth projective variety in $\mathbb{P}^n$, over the field of complex numbers or an algebraically closed field of characteristic $0$.

EDIT (After Stefano's response): Assume $\dim X < n-1$, $\deg X > 1$ and $X$ is nondegenerate.

In particular, $X$ is not a hypersurface. Let $P\in X$ be a general point, fix a general hyperplane in $\mathbb{P}^n$ and consider the map $\varphi_P:X\setminus \{P\}\to \mathbb{P}^{n-1}$ obtained by "projecting from $P$". Let $Y$ be the closure of the image $\varphi_P(X)$ in $\mathbb{P}^{n-1}$.

(1) For general $P$, is the map $\varphi_P:X\setminus\{P\}\to Y$ going to be quasi-finite?,

(2) For general $P$, is the map $\varphi_P:X\setminus\{P\}\to Y$ going to be birational?

I am guessing that these are true. I could not find a reference for these two statements in standard texts like Harris, Hartshorne or Shararevich . Can someone please point out a reference for these. If someone could sketch a proof or explain a counterexample then that would be more helpful. Thanks.

$\endgroup$
0
$\begingroup$

I think that to have (1) or (2) you need to put extra assumptions (I am not sure whether it will work even then).

One condition that you need is that $X$ is non-degenerate, i.e. it is not contained in any hyperplane. Indeed you can consider a plane cubic $E$ as a curve in $\mathbb{P}^3$. Then, the projection will be $2:1$, since every line will meet two residual points. Thus, the map is not birational.

Similarly, you can consider a quadric surface $Q \subset \mathbb{P}^3$ as embedded in $\mathbb{P}^4$. Then, through $P$ there are exactly two rulings ($Q$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$), and the map contracts each ruling to a point. Thus, the map is not quasi-finite.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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