There are answers to this elsewhere, but my question is in the context of a particular problem, that being 1.4.21 from Garrity's Algebraic Geometry: A Problem Solving Approach.
Exercise 1.4.21. Once we have homogenized an equation, the original variables $x$ and $y$ are no more important than the variable $z.$ Suppose we regard $x$ and $z$ as the original variables in our homogenized equation. Then the image of the $xz$-plane in $\mathbb{P}^2$ would be $\{(x:y:z) \in \mathbb{P}^2 : y = 1\}.$
(1) Homogenize the equations for the parallel lines $y=x$ and $y = x + 2.$
The homogenizations are $y = x$ and $y = x+ 2z.$
(2) Now regard $x$ and $z$ as the original variables and set $y = 1$ to sketch the image of the lines in the $xz$-plane.
This gives us $x = 1$ and $z = -\frac{1}{2}x + \frac{1}{2}.$ The graph looks like the following: https://www.desmos.com/calculator/4ncyzir9kg
(3) Explain why the lines in part (2) meet at the $x$-axis
We can see in the picture that the lines intersect at the x-axis.
I know that the original lines intersect at the line at infinity. I can even show this algebraically. If we set the equations $x = 1$ and $z = -\frac{1}{2}x + \frac{1}{2}$ equal to each other, then we get that $z = 0.$ We already know that $x = 1$ and $y = 1,$ so the intersection point is $(1:1:0),$ which is the line at infinity. My question is how parts (2) and (3) are supposed to show this. I'm looking for some geometric intuition as to why intersecting at the $x$-axis means that the original lines intersect at the line at infinity.
Best Answer
Taking a “top-down” approach, there is no distinguished line at infinity in the pure projective plane, nor is there a notion of parallel lines. Any line can be singled out as the line at infinity, often implicitly by the coordinate system that you impose on $\mathbb P^2$, which then in turn defines which lines are parallel. (In doing so you have imposed an affine geometry on the projective plane.) That’s more or less what the text is trying to communicate by switching the line at infinity from $0:0:1$ to $0:1:0$.
In the “bottom-up” model of $\mathbb P^2$ being used by your text, points in $\mathbb P^2$ correspond to lines through the origin in $\mathbb R^3$ (strictly speaking, in $\mathbb R^3\setminus\{0\}$), and lines in $\mathbb P^2$ correspond to planes through the origin in $\mathbb R^3$. Obviously, any two such distinct planes intersect in a line through the origin—a point in $\mathbb P^2$. The correspondence between lines/planes in $\mathbb R^3$ and point/lines, respectively, in $\mathbb P^2$ is established by choosing a plane in $\mathbb R^3$ that doesn’t go through the origin—conceptually, we embed $\mathbb P^2$ into $\mathbb R^3$ as that plane. The intersections of the lines/planes through the $\mathbb R^3$ origin are the corresponding projective points/lines; the parallel plane through the origin maps to the projective line at infinity and the lines that lie on this plane are the projective points at infinity.
The standard coordinate system used in the text corresponds to choosing the plane $z=1$ as the embedded projective plane. If two projective lines intersect at infinity, then the intersection of their corresponding planes in $\mathbb R^3$ lies on the $x$-$y$ plane. In this exercise, you are then asked to switch to $y=1$ as the embedded projective plane. If two lines on this plane intersect on the $x$-axis, then the line in $\mathbb R^3$ that represents this point lies entirely within the $x$-$y$ plane, but per the preceding that means that the projective lines intersected at infinity in the original coordinate system that has $0:0:1$ as the line at infinity. On the other hand, they clearly don’t intersect at infinity in the second embedding. This change of embedding is equivalent to applying a projective transformation to $\mathbb P^2$ and illustrates that parallelism of lines is not a projective invariant—it’s not a property that’s preserved by projective transformations.
Here are the planes (in red) that correspond to the two lines in the problem and their line of intersection (orange). This line lies entirely within the $x$-$y$ plane (gray).
In the original embedding, the two lines represented by the red planes are their intersection with the plane $z=1$:
Here, the plane $z=0$ corresponds to the line at infinity. The red planes’ intersection lies on this plane, so they are parallel and intersect at a point at infinity.
In the second embedding, we instead intersect the two red planes with $y=1$:
In this embedding, the plane $y=0$ (not shown) corresponds to the line at infinity, and the lines represented by the red planes are no longer parallel. The orange line intersects $y=1$ at a single point—the point of intersection of the two lines on the embedded plane. The black lines are not parallel in this embedding, but their intersection does lie on the embedded plane’s $x$-axis, which is its intersection with the $x$-$y$ plane in $\mathbb R^3$, the original line at infinity.