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Why is a point at infinity the point where the last element in a homogeneous point vector is 0?

Here's the context:

enter image description here

I'm trying to imagine the continuity as the third coordinate in the homogeneous line approaches 0. It's clear that when it equals 0, the two homogeneous lines share the same normal vector, so we consider them to be equal. But what if the third coordinate equals an arbitrarily small value? What do these lines and their normals look like?

enter image description here

Also, in a lecture I'm watching, the explanation for why distant objects in this image appear small is that we divide by $Z$ when converting from homogeneous points to Euclidian points. Why is this so? It seems that $Z$ represents distance from the camera, and for a very large $Z$ (not a very small $Z$), points become very close to each other. It's strange that points equal each other when $Z$ is 0. How can I visualize $Z$?

enter image description here

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  • $\begingroup$ I would suggest you to see this vide. It is the best I have found on this topic. $\endgroup$
    – Babu
    Commented May 23, 2022 at 11:27

2 Answers 2

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You could take this as a definition of "points at infinity"; to wit, that's exactly what your quoted exposition is doing.

You can visualize why this is a reasonable name by considering the mapping back from homogeneous vectors to affine points:

$$ (X : Y : Z) \mapsto \left( \frac{X}{Z} , \frac{Y}{Z} \right) $$

when $Z=0$, you have a division by zero, thus things blow up to infinity.

If you've seen a geometric construction of projective space, defining the points at infinity in a way that every class of parallel lines intersects at a unique point 'at infinity', then we can see this more directly. In affine coordinates, the equation of a line passing through the origin is given by

$$ ax + by = 0 $$

the homogenization of this equation is

$$ aX + bY = 0$$

and so the set of all points on this line in homogeneous coordinates is $(b : -a : t)$ where $t$ can be anything. The only point on the line that isn't an ordinary point is the $Z = 0$ case.

If you take the limit as $t \to 0$, you can see that the point is zipping off to infinity. This might be easier to see in affine coordinates, where for $t \neq 0$ the cooresponding affine point is $\left(\frac{b}{t}, -\frac{a}{t} \right)$.

If you took another parallel line with the same slope; e.g. $ax + by = 1$, the homogenization is $aX + bY = Z$ and the general solution is $(b : t-a : bt)$. Again at $t = 0$ this point becomes $(b : -a : 0)$, the same point we got in the previous example.

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  • $\begingroup$ What does it mean for $Z$ to be less than 1? When Z is 1, the representation of the point in 2D space is the same as in homogeneous space, so I thought the xy plane intersects with the plane $Z=0$. The algebra makes sense, but shifting the $Z$ plane back by 1 and gives me no geometric intuition on why parallel lines are equal when $Z=0$ (although the math does check out). This PDF has some good diagrams that perhaps you can use to help answer: campar.in.tum.de/twiki/pub/Chair/TeachingWs05ComputerVision/… $\endgroup$ Commented Mar 26, 2014 at 14:55
  • $\begingroup$ @Rose: Projective geometry originally came from the geometric observation of point-line duality: there were a number of theorems that people were discovering that, if you swapped the words "point" and "line" and fiddled with the details, you got another theorem. The ultimate desire for projective geometry for there to be no details to fiddle with: the duality was exactly right. In particular, we need the axioms of geometry to be dual: because "through any two distinct points there is a unique line" we also want "lying on any two distinct lines there is a unique point" ... $\endgroup$
    – user14972
    Commented Mar 26, 2014 at 15:28
  • $\begingroup$ ... thus, a pair of parallel lines must have a point of intersection when we move to projective geometry. And sine this point isn't an 'ordinary' point, that intersection has to be an ideal point. $\endgroup$
    – user14972
    Commented Mar 26, 2014 at 15:29
  • $\begingroup$ Is there continuity as $Z$ approaches 0? $\endgroup$ Commented Mar 26, 2014 at 15:32
  • $\begingroup$ @Rose: Yes. P.S. I reread your previous comment, and I have another observation: your problem might be that you're thinking of an "xy plane" and a "z plane" -- you're trying to imagine three-dimensional space. But we're working in a two-dimensional projective plane: while we have three coordinates, there is one degree of redundancy.... $\endgroup$
    – user14972
    Commented Mar 26, 2014 at 15:35
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$\newcommand{\R}{\mathbf{R}}\renewcommand{\P}{\mathbf{P}}$Note that in the projective plane $\P^{2}$, there's no intrinsic notion of points lying at infinity. Instead, a point of $\P^{2}$ (i.e., a line through the origin of $\R^{3}$) may (or may not) lie at infinity with respect to a choice of "affine coordinate chart" (or "dehomogenization").

As your book says, the set of points at infinity with respect to any particular affine chart is a projective line. The points on this "line at infinity" correspond geometrically to directions in the affine chart. Algebraically, points in the chart $Z \neq 0$ are represented uniquely by dehomogenized triples: $$ [X : Y : Z] \leftrightarrow [x : y : 1] = (x, y),\quad x = X/Z,\ y = Y/Z. $$ A point at infinity in this chart is (by definition) a point of $\P^{2}$ not represented by these affine coordinates, namely, a triple with $Z = 0$, a.k.a. a line through the origin of $\R^{3}$ in the $(X, Y)$-plane. Starting at the "origin" $(0, 0)$ of the affine chart and traveling along the line through $(x, y) \neq (0, 0)$ is tantamount to looking at points of $\P^{2}$ represented by $$ (tx, ty) = [tx : ty : 1] = [x : y : \tfrac{1}{t}]; $$ as $|t| \to \infty$, the point $(tx, ty)$ therefore approaches $[x : y : 0]$, the "point at infinity along the direction $(x, y)$".

Regarding your question about distance and apparent size, it looks to me that you're trying to reconcile two different notions of convergence. Specifically, suppose $[X : Y : Z]$ is a point with $Z \neq 0$ (but free to vary) and $(X, Y) \neq (0, 0)$ is fixed. In the affine chart with $(x, y) = (X/Z, Y/Z)$:

  1. If $Z \to 0$, then $(x, y)$ approaches $[x : y : 0]$ in $\P^{2}$; the same is true of every point on the affine line through $(0, 0)$ and $(x, y)$ (other than $(0, 0)$ itself). This recapitulates the preceding observation that points at infinity in an affine chart correspond to affine directions in that chart.

  2. If $|Z| \to \infty$ (the object moves away from the camera, staying parallel to the viewing axis $[0 : 0 : 1]$), then $(x, y) \to (0, 0)$. Loosely, "distant objects appear to be close to the origin". This recapitulates your video lecture ("...divide by $Z$..."), and explains why parallel railroad tracks "meet at infinite distance". The terminology is potentially confusing: This meeting point at infinite distance along the tracks does not lie in the line at infinity in screen coordinates.

Generally, in the chart $Z \neq 0$, the line $\ell$ in $\R^{3}$ defined parametrically by $[X_{0} + tA : Y_{0} + tB : Z_{0} + tC]$ (with $C \neq 0$, and not passing through the origin of $\R^{3}$) projects to the affine line $$ \left(\frac{X_{0} + tA}{Z_{0} + tC}, \frac{Y_{0} + tB}{Z_{0} + tC}\right). $$ As $|t| \to \infty$, this point approaches $(A/C, B/C)$, the common vanishing point for lines with direction vector $(A, B, C)$. Again, this vanishing point lies at "infinite distance" along $\ell$, but represents a (finite) location in the affine $(x, y)$ coordinates.

It may be a helpful exercise to sketch all this in $\R^{3}$, "unwrapping" $\P^{2}$ to the unit sphere.

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