Questions about the projective plane

Question

On Wikipedia it is stated that

points of the form $[x:y:1]$ are the usual real plane and

points of the form $[x:y:0] $ are the line at infinity.

But this choice $z=0$ seems arbitrary to me. The projective space seems pretty symmetric so one might as well say $[0:y:z]$ or $[x:0:z]$ are the line at infinity.

What am I missing? Why is a point at infinity iff $z=0$?

The choice $z=1$ for the real plane seems equally arbitrary. I expect everything but the line at infinity to correspond to the usual real plane.

Again:

What am I missing? Why does $[x:y:z]$ (with $x,y,z\neq 0$) not correspond to a point on the real plane but $[x:y:1]$ does?

Answer 1

It is indeed arbitrary to choose $z=0$ to be the line at infinity. It would be just as reasonable to choose $y=0$, or $x=0$, or really any other line in the projective plane. What you have observed is that the symmetries of the projective plane do not preserve the line at infinity: that is, the "line at infinity" is not an intrinsically defined subset of the projective plane, but merely an arbitrary choice. It is conventional to choose $z=0$.

As for the points in the usual plane being $[x:y:1]$, here you really are missing something. Remember that by definition, $[x:y:z]=[tx:ty:tz]$ for any nonzero $t$. This means that if you have a point $[x:y:z]$ with $z\neq0$, then it is equal to the point $[x/z:y/z:1]$. So every point with $z\neq 0$ can be written a form where its last coordinate is $1$. In fact, it can be written uniquely in this form, since $1/z$ is the only choice of $t$ to multiply that will make the last coordinate $1$.

Answer 2

The projective plane does indeed have all of the symmetry that you impute to it. But the particular coordinate system does not: it makes it possible, by setting $z=1$, to find that one special plane within it is the "Cartesian" $(x,y)$-plane. In 8th grade you learned about the parabola in the $(x,y)$-plane whose equation is $y=x^2$. Now you can "homogenize" that equation, making every term a second, degree term, by writing $zy=x^2$. Notice that if $(x,y,z)$ satisfies the equation $zy=x^2$, then $(cx,cy,cz)$ does as well (for every $c\ne0$), and therefore this homogeneous equation $zy=x^2$ identifies a curve in the projective plane. In the real projective plane, all ellipses are created equal, and every ellipse becomes a parabola in the affine plane if the line chosen to be at infinity is a line passing through some point that is one the ellipse. And so $zy=z^2$ is one equation of an ellipse.

In a genuine projective plane (over a field, such as $\mathbb R$), all lines are created equal. Choosing any particular line to become the line at infinity leaves an affine plane that excludes that line. But that doesn't mean a particular choice of coordinate system treats all lines equally.