# Generic linear projection with $\text{Codim}\ge 2$ is birational (geometric proof)

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?

• 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.) Commented Jul 1, 2020 at 23:42
• @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. Commented Jul 2, 2020 at 0:22
• 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 $d$. Commented Jul 2, 2020 at 9:08
• Dear @Sasha: $X$ has codimension at least two in my assumption. (If it were a hypersurface, I agree with what you said.) Commented Jul 2, 2020 at 13:22

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.
• 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 $X$? Commented Jul 3, 2020 at 4:11
• @AGlearner: Yes, $\eta(Y)$ is the secant variety. Commented Jul 3, 2020 at 8:08