3. Points at infinity

So far we have discussed homogeneous coordinates for ordinary points such as $(3,4)$. Because these coordinates result from multiplying an extended vector such as $(3,4,1)_h$ by a nonzero scalar, the third number is never 0.

On the other hand, applying a projective transformation could very well result in a triple ending in 0. In Example above, for instance, we could compute

$T(3,-2) = pt (3,-2,1)_h \left[\begin{array}{ccc} 1&0&0\\ 0&1& {\frac 1 2}\\ 0&0&1 \end{array}\right]$ $=$ $pt (3,-2,0)_h$,

but it's not clear what $pt (3,-2,0)_h$ means, since there's no such point in the usual sense.

Can a meaning be attached to a triple ending in 0? The answer is yes (except for the hopeless case $(0,0,0)$ ). Let's sneak up on $pt (3,-2,0)_h$ by looking at the sequence of points

$pt ( 3, -2, 1)_h$, $pt ( 3, -2, {\frac{1} {2}})_h$, $pt ( 3, -2, {\frac{1} {3}})_h$, $pt ( 3, -2, {\frac{1} {4}})_h$, $pt ( 3, -2, {\frac{1} {5}})_h$,

$pt ( 3, -2, {\frac{1} {6}})_h$, $\dots$

Algebraically, these triples have limit $(3, -2, 0)$. To picture them graphically, just represent them in Cartesian coordinates. For example, $pt ( 3, -2, {\frac{1} {2}})_h$ $=$ $pt (6, -4, 1)_h$ $=$ $(6,-4)$. We get the points $(3,-2)$, $(6,-4)$, $(9,-6)$, $(12,-8)$, $(15,-10)$, $(18,-12)$, and so on:

book/04dir/limit.eps

As you see, this sequence of points is headed off the real plane, in the direction given by the vector v$= (3,-2)$. We get the following idea:

The homogeneous coordinates $(a,b,0)_h$ represent a point at infinity along the line through the origin with direction vector $(a,b)$.

Of course, a point at infinity is in one sense an imaginary invention, but it does have a concrete reality in that it is representable by numbers and corresponds to a line through the origin. For these reasons, we can talk about points at infinity and be sure that we will not run into trouble.

Let us call the usual points of the Cartesian plane ordinary points, in contrast to the points at infinity.

Several interesting observations:

Observation 3.1 . Because $(3,-2,0)_h$ and $(-3,2,0)_h$ differ by a scalar factor, namely $-1$, they represent the same point at infinity. Thus the point at infinity corresponds more to the line itself than to one direction on the line. You can picture the line as somehow ``wrapping around'' at its point at infinity.

Observation 3.2 . Every line with direction $(3,-2)$ goes towards the same point at infinity. For example, the line $P(t) = (4,5) + t(3,-2)$ has points $(4+3t,5-2t)$, which in homogeneous coordinates are $pt (4+3t,5-2t,1)_h$ $=$ $pt ( {\frac{4} t} + 3, {\frac{5} t} - 2, {\frac{1} t})$. As $t \rightarrow \infty$, this triple has limit $(3,-2,0)_h$.

Thus a point at infinity corresponds to a family of parallel lines (i.e., the set of all lines parallel to a given line). In some sense, the parallel lines meet at that point at infinity.

Problem 3.3 . What point at infinity lies on the line through $(2,3)$ and $(3,5)$?

Solution. The vector between the two points is $(1,2)$. This vector gives the direction of the family of all lines parallel to the given line. Therefore the given line goes through the point $pt (1,2,0)_h$ at infinity.

Observation 3.4 . A projective transformation can take points at infinity to ordinary points, and vice-versa. For example, in going from the real football field to the television screen, the point at infinity where the yard-lines ``meet'' is mapped to the ordinary point where their images meet, off the top of the page. Also, the ordinary point off the field where the two lines of the band would meet (if extended), is mapped to a point at infinity (because the images of the two lines are parallel).