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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?

self-learner
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  • Remember that in projective space, you can scale your points $[x,y,z]$ arbitrarily by non-zero real numbers. In other words, for all $\lambda \in \mathbb{R}$, $[x,y,z] = [\lambda x, \lambda y, \lambda z]$. You can cover the projective line by the three open sets where you let $x$, then $y$, then $z$ not be zero. By the irrelevance of scaling described above, you can rescale the coordinates such that there's a 1 the slot, i.e $[x/z,y/z,1]$ and likewise for the other two open sets. This gives you something like $\mathbb{R}^{2}$ and when $z=0$, you get a point at infinity. – Benighted Jul 08 '16 at 03:06
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    You are right, the standard homogeneous coordinates approach is one of infinitely many ways of embedding the ordinary plane in the real projective plane. Any line in the projective plane can be chosen as "the line at $\infty$." – André Nicolas Jul 08 '16 at 03:09
  • @AndréNicolas Thank you for your comment. And is it correct to say that the points $[x:y:z]$ with $z\neq 0$ is the real plane? I think it should be because one can scale. – self-learner Jul 08 '16 at 03:39
  • Yes, sort of. It is really the equivalence class of $(x,y,z)$, where $z\ne 0$, and two triples, not all $0$, are equivalent if one is a constant times the other. You can think of it as a plane through the origin other than the $x$-$y$ plane. – André Nicolas Jul 08 '16 at 03:46

2 Answers2

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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 $1$.

Eric Wofsey
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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.

Michael Hardy
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  • An aside that doesn't affect your main point, there are lots of genuine projective planes in which not all lines are created equal. In fact, it is conjectured that among finite projective planes, this is only true of $PG(2,q)$. – xxxxxxxxx Jul 13 '16 at 14:02
  • @MorganRodgers : I had in mind a projective plane over a field and especially the sort of projective plane contemplated in the posted question. $\qquad$ – Michael Hardy Jul 13 '16 at 15:21