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Consider the quadric surface $X = \{ xy = zw \} \subset \mathbb{P}^3$ and pick a point $x \in X$. I think it is true that if we think of $\mathbb{P}^2$ as the space of lines through $x$ in $\mathbb{P}^3$, then the morphism $X \setminus \{ x \} \to \mathbb{P}^2$ which sends $y \mapsto \overline{xy}$ represents a birational map $X \to \mathbb{P}^2$. But I do not understand the geometry of $X$ well enough to prove this. Certainly this morphism fails to be injective along the two obvious lines in $X$ through $x$, but how do I see that the map is an isomorphism elsewhere? I would like to avoid computing in coordinates if at all possible.

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Nice question:+1 – Georges Elencwajg May 14 '12 at 19:20
up vote 3 down vote accepted

The basic idea is that the "inverse map" is given by sending $\ell \in \mathbb{P}^2$ (identify the points in $\mathbb{P}^2$ with lines through $x$) to the point $y$ where $X \cap \ell = \{x, y\}$. (We are using that $X$ has degree 2, which means that it intersects a general line in 2 points.) It's pretty clear that this map is inverse to the one you described, wherever things are well-defined. It also should be clear that the sets of points in $X$ and $\mathbb{P}^2$ where the maps are not well-defined are proper closed subsets. Lastly, you need to check that these are actually morphisms where they are defined, and I'm afraid this step, by definition, requires a certain amount of coordinate computation to verify.

A side remark to help you understand the geometry of $X$ is that you can view it as the isomorphic image of $\mathbb{P}^1 \times \mathbb{P}^1$ under the Segre map $((a:b),(c:d))\mapsto (ac:bd:ad:bc)$ to $\mathbb{P}^3$. I chose a weird ordering for the products of the variables so that the equation $xy = zw$ would be satisfied, assuming you order your coordinates on $\mathbb{P}^3$ "alphabetically" as $(x:y:z:w)$. In particular, this shows that through every point of $X$ are exactly 2 lines in $\mathbb{P}^3$ contained in $X$. Indeed, the two copies of $\mathbb{P}^1$ give two separate rulings on $X$. Shafarevich's book has a nice discussion of this surface; you might also look at Igor Dolgachev's notes on Classical Algebraic Geometry.

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Nice answer:+1 ${}$ – Georges Elencwajg May 14 '12 at 19:21
Thanks: can you briefly explain how you know that exactly two lines pass through each point? Why not more? – Justin Campbell May 14 '12 at 19:32
Sure: suppose $p \in X$ is the image of $(q_1,q_2) \in \mathbb{P}^1 \times \mathbb{P}^1$ under the Segre map. Then the two lines through $p$ are the images of $q_1 \times \mathbb{P}^1$ and $\mathbb{P}^1 \times q_2$. – Michael Joyce May 14 '12 at 19:52
@MichaelJoyce: I see these two lines, but how do you know there are no other lines through the point? This is what I do not understand. – Justin Campbell May 14 '12 at 21:16
There are two (interrelated) ways I can think of to see this. First, you can do a calculation with coordinates; that is, write down a general line in $\mathbb{P}^3$ and compute its intersection with the image of the Segre map. Second, if you have a linear form $L$ on $\mathbb{P}^3$ that vanishes on the image of the Segre map $\iota$, then $L \circ \iota$ must have multidegree $(1,0)$ or $(0,1)$ on $\mathbb{P}^1 \times \mathbb{P}^1$. But these give you precisely the lines of the form mentioned above, and it is easy to check that only the two already mentioned pass through $p$. – Michael Joyce May 14 '12 at 23:53

As a concrete example of this problem, consider the projection of $X$ through the point $[0:0:0:1]$ onto the plane $w=0$. Show this is a birational map (consider the Segre embedding above, where is it not defined?) and agrees with your construction above. I know this uses coordinates, but it is actually not too messy.

Edit: Let me elaborate a little bit. If you consider the affine patch where $w=1$ we get the equation $z=xy.$ The point mentioned above maps to $(0,0,0).$ This only contains the two families of lines mentioned in the previous post, so away from these two lines at the origin (namely the $x,y$ axes), the map to $\mathbb{P}^2$ is an isomorpism. Namely, send a point in $X$ to the the approriate line, and send a line to the associate intersection with $X.$

This picture helped me:

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