Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
Nice question:+1 –  Georges Elencwajg May 14 '12 at 19:20

2 Answers 2

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.

share|cite|improve this answer
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:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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