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Let $X\subseteq \mathbb P^N$ be a smooth variety and $p$ a general point $\mathbb P^N\setminus X$. Consider the map $$\pi:X\to \mathbb P^{N-1}$$

as the restriction of linear projection $\mathbb P^N\setminus \{p\}\to \mathbb P^{N-1}$. Then $\pi$ is a finite morphism (proper and quasi-finite) onto its image. Let's assume $\dim X\le N-2$, in other words, $\pi(X)$ is not the whole space, then is $\pi:X\to \pi(X)$ a birational map?

Edited: As KReiser pointed it out, it is an exercise in Hartshorne (which I didn't realize), and the standard proof is to show $X$ and $\pi(X)$ have the same function field using the standard tool in commutative algebra. There are already complete solutions linked in KReiser's comment.

However, as a complex geometer, I'm used to thinking this problem geometrically, for example, I can relax the condition to $X\subset \mathbb C^N$ an analytic subvariety and ask if a generic linear projection is bimeromorphic. In other words, to me this problem is equivalent to show that for a generic point $q\in \pi(X)$, the line through $\overline{pq}$ intersect $X$ at only one point. Is there a geometric proof of this fact?

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  • $\begingroup$ The answer is yes, and this is Hartshorne exercise I.4.9. See solutions on this website here, here, etc. (If you have a favorite, let me know so I can mark this as a duplicate. If these links don't answer your questions, please add an edit with the additional information you require.) $\endgroup$
    – KReiser
    Commented Jul 1, 2020 at 23:42
  • $\begingroup$ @KReiser: Thanks for pointing that out, that's already great solutions, but I think I'm still looking for a geometric solution, so I edited my post. $\endgroup$
    – AG learner
    Commented Jul 2, 2020 at 0:22
  • $\begingroup$ If $\dim(X) = N - 2$, so it is a hypersurface, and $d := \deg(X) > 1$, so it is not a hyperplane, then the morphism is NOT birational; indeed its degree is $d$. $\endgroup$
    – Sasha
    Commented Jul 2, 2020 at 9:08
  • $\begingroup$ Dear @Sasha: $X$ has codimension at least two in my assumption. (If it were a hypersurface, I agree with what you said.) $\endgroup$
    – AG learner
    Commented Jul 2, 2020 at 13:22

1 Answer 1

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Assume $\dim(X) = n$. The map $$ X \times X \to Gr(2,N+1), \qquad (x_1,x_2) \mapsto \langle x_1, x_2 \rangle, $$ that takes a pair of points to the line spanned by them is not defined on the diagonal, but after its blowup extends to a regular morphism $$ Bl_{\Delta}(X \times X) \to Gr(2,N+1). $$ Let $$ Y = Bl_{\Delta}(X \times X) \times_{Gr(2,N+1)} Fl(1,2;N+1). $$ This variety, roughly speaking, parameterizes triples $(x_1,x_2,y)$, where $x_1,x_2 \in X$ and $y$ is on the line in $\mathbb{P}^N$ spanned by $x_1,x_2$. Note that $\dim(Y) = 2n + 1$. Consider the projection $$ \eta \colon Y \to \mathbb{P}^N, \qquad (x_1,x_2,y) \mapsto y. $$ Note that for general point $p \in \mathbb{P}^N$ we have $$ \dim(\eta^{-1}(p)) \le 2n + 1 - N = n - (N - n - 1) < n. $$ Therefore, the image in $X$ of $\eta^{-1}(p)$ has positive codimension. This means that if you project from $p$ then for general point $x_1 \in X$ the point $p$ does not lie on a secant line to $X$ passing through $x_1$. This implies that the projection from $p$ is birational.

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  • $\begingroup$ This is exactly the proof I'm looking for. Thank you for sharing the idea! By the way, is it true that $\eta(Y)$ is just the secant variety of $X$? $\endgroup$
    – AG learner
    Commented Jul 3, 2020 at 4:11
  • $\begingroup$ @AGlearner: Yes, $\eta(Y)$ is the secant variety. $\endgroup$
    – Sasha
    Commented Jul 3, 2020 at 8:08

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