Why is the homogeneous line through all points at infinity (1:0:0) and not (0:0:0)?

Question

So I just had a geometry lecture that introduced me to homogenous coordinates. To be clear with notation let me recap: Homogenous coordinates in $\mathbb R^n$ space are described as $$(x_0:x_1: ... : x_n)$$ where $x_0$ is the factor by which to divide to get cartesian coordinates: $$(\dfrac{x_1}{x_0}:\dfrac{x_2}{x_0}: ... : \dfrac{x_n}{x_0})$$ next up we learned that the point at both ends of of a line in 2 dimensions - and each line parallel to it - is defined as $$(0:x_1:x_2)$$ Now here's my question: After hearing this I concluded that the line through all points at infinity must be $(0:0:0)$ but instead we got told that the line homogenous coordinates for that line are $(1:0:0)$ and that the point at $(0:0:0)$ does not exist. What is the reasoning behind this? How does this follow from the definiton of other points at infinity?

Answer 1

Consider the $x_0=1$ plane $\Sigma$ in the $x_0,x_1,x_2$ coordinate space.
Every line through the origin meets $\Sigma$ in exactly one point, except for the ones that parallel to $\Sigma$ (having direction vector of the form $(0,x_1,x_2)$), but they will just correspond to the points at infinity.

The planes in the space through the origin intersect $\Sigma$ in a line, except the $x_1,x_2$ plane, (that is the plane $x_0=0$) and that will correspond to the line at infinity.
Note that the planes in the space - hence the lines in the projective plane - are determined by their normal vector.
Consequently, the line at infinity of $\Sigma$ is determined by $(1,0,0)$, also meaning that $(x_0:x_1:x_2)$ is a point on that line iff $(1,0,0)\pmatrix{x_0\\ x_1\\ x_2}=0$, i.e. $x_0=0$.

Answer 2

I'll try to do some clarification, hopefully you question is somehow covered in this.

• You seem to think that cartesian and homogenous coordinates are just to different coordinate-systems for the same space. Thats not true. Cartesian coordinates label elements of the euclidean space $\mathbb{R}^n$ (which is an ordinary vector space), while homogenous coordinates label elements of the projective space $\mathbb{PR}^n$ (which is not a vector space b.t.w.).
• The projective space $\mathbb{PR}^n$ can be defined as the set of rays through the origin in $\mathbb{R}^{n+1}$. I.e. as a set of equivalence classes $$\mathbb{PR}^n=\left(\mathbb{R}^{n+1}\setminus\{0\}\right)/ \simeq$$ where $\simeq$ is the equivalence relation $$x\simeq y\iff \exists \lambda\in\mathbb{R}: \lambda x=y$$ This means that for example $(1:2:3)$ and $(2:4:6)$ label the same point of $\mathbb{PR}^2$. It also means that $(0:0:0)$ does not exist because that cannot correspond to a ray.
• Alternatively, you can think of $\mathbb{PR}^2$ as the surface of a sphere as in this picture from Wikipedia https://en.wikipedia.org/wiki/Projective_space#/media/File:Projective_plane2.png The straight blue lines are in the euclidean space, and they correspond in the projective space to the blue (half-)circles. In the projective space, these half-circles have natural "end-points" at the "equator" (which are points of the form $(0:x_1:x_2)$). These "end-points" do not correspond to any point in the euclidean space.
• The set of all "end-points" form the "euqator" (green circle in the image), which is in fact a line in the projective space. You can write this line as $$\{(0:x_1:x_2)\ | \ x_1,x_2\in\mathbb{R}, x_1\neq 0 \vee x_2\neq 0 \}$$ So neither $(1:0:0)$ (which is just the "north-pole" in the picture) nor $(0:0:0)$ (which does not exist at all) is part of the "infinity-line". I don't now what your teacher wanted to say at that point.

Note 1: You should use the ':' notation only for homogenous coordinates in $\mathbb{PR}^n$, not for cartesian coordinates in $\mathbb{R}^n$. So your mapping should be written as $$(x_0:x_1:...:x_n)\mapsto (\frac{x_1}{x_0},...,\frac{x_n}{x_0})$$

Note 2: Calling the points of the form $(0:x_1:x_2)$ "end-points" might actually be misleading: In the projective space there is nothing special about these points, and the circles on the sphere actually go all the way around without any distinguished points at all. But note that to "points" at opposite sites of the sphere are actually the same projective point, because $(x_0:x_1:x_2)=(-x_0:-x_1:-x_2)$, thats why the lower half is not not drawn in the picture because it would be the same points as the upper half.

Answer 3

A point $\mathbf p$ lies on the line $\mathbf l$ iff $\mathbf l\cdot\mathbf p = 0$. If the line at infinity were represented by $(0:0:0)$, then every point would lie on this line, which is not at all what we want. The representation of the line that passes through only points with coordinates of the form $\mathbf p = (0:x_2:x_3)$ must satisfy $\mathbf l\cdot(0:x_2:x_3)$ for all $x_2$ and $x_3$, which means that its second and third components must be $0$. The remaining component must be nonzero, which gives $\mathbf l = (1:0:0)$.

More generally, the zero vector is never a valid homogeneous coordinate representation of a point, line, plane, &c, although it can appear as the result of a computation. For example, if $L$ is the Plücker matrix of a line on $\mathbb{RP}^3$, its intersection with a plane $\mathbf\pi$ is given by $L\mathbf\pi$. If this product is the zero vector, that means that the line lies on $\mathbf\pi$.