Jtdylktuy

I'm working through an old textbook called Algebraic Projective Geometry, by Semple and Kneebone.

Early in the text, the authors write:

When we introduced complex points (on p. 12) we explained that they are to be regarded as ideal points adjoined to the euclidean plane; and the same is now true of points at infinity. These too are ideal points, but since they have not been introduced by way of coordinate representation it is not immediately obvious how they can be handled algebraically. This can, in fact, be managed quite easily, as we shall now show, by making the coordinates homogeneous.

When a pair of axes has been chosen, all actual (i.e. non-ideal points of the plane admit of unique and unexceptional representation by pairs of real numbers (X, Y). Two lines, whose equations are

aX + bY + c = 0 and a'X + b'Y + c' = 0,

meet, if they are not parallel, in the actual point whose coordinates are given by

X : Y : 1 = bc' - b'c : ca' - c'a : ab' - a'b.

Now whether or not the lines are parallel, supposing only that they are distinct, the ratios

bc' - b'c : ca' - c'a : ab' - a'b

are always determinate, in the sense that at least one of the quantities concerned is different from zero. Our object, then, must be to represent points by such sets of ratios rather than by pairs of numbers which are liable to become infinite.

I don't understand how the ratios are always determinate. I can't see that at least one of the quantities concerned is different from zero. Could you show me?

Any further elaboration on the motivation and purpose of homogeneous coordinates would be appreciated (I know this is the type of vauge/big question that this StackExchange may not like, but I do computer vision, and while I have a vague notion of the motivation and purpose of homogeneous coordinates, another attempt at explaining them to me would never hurt...)

Think of this:

$$bc'-b'c=0iff bc'=b'cimpliedby b:b'=c:c'$$

So essentially a zero entry in the equation for the point of intersection means that the ratio between the coefficients for one and that for the other line are equal. I said “essentially” because there might be zeros in there, making the rightmost implication not quite an equivalence. I'll cover that in a moment, after explaining what I'm aiming for.

Assume that all three terms in the result are zero. Then you get three such equations, which taken together express $a:a'=b:b'=c:c'$. So the ratio between the coefficients is the same, which means one line equation is just a multiple of the other. If the equations are multiples of one another, they describe the same line. This is excluded by assuming distinct lines.

But what if one of the denominators at the right is zero? Then you can't divide by it in the normal sense. Assume without loss of generality that $b'$ is zero. Then the middle equation becomes $bc'=0$ so either $b$ or $c'$ have to be zero as well. And since you have a similar equation $ab'=a'b$ from another zero in the result, you also need $b$ or $a'$ to be zero. If you take $b=0$, the above argument still holds: one equation is a multiple of the other, even if one of the coefficients happens to be zero. If instead you assume $bneq 0$, then you have to take $a'=0$ and $c'=0$, summarized as $a'=b'=c'=0$, and the second equation becomes null, not representing a line any more.

So if the two inputs are actually lines (i.e. have at least one non-zero coefficient, which will describe the line at infinity for $a=b=0, cneq0$), and if the lines are distinct (one line equation is not a multiple of the other), then the point of intersection computed as described will have at least one non-zero coordinate as well.

Viewing the homogeneous coordinates of a point as elements of $mathbb R^3$, we can write $mathbf x = (X,Y,1)^T$ and $mathbf l = (a,b,c)$ so that the equation $aX+bY+c=0$ of a line becomes $mathbf l^Tmathbf x = 0$. This is just the dot product of the two vectors, and in $mathbb R^3$ is the equation of a plane through the origin with normal $mathbf l$. If two such planes intersect in a line, then this line is orthogonal to both normals, so its direction vector is the cross product of the two normals: $$mathbf ltimesmathbf m = (bc'-b'c,ca'-c'a,ab'-a'b)^T.$$ The two normals are nonzero, so the only way for this cross product to vanish is for $mathbf l$ and $mathbf m$ to be parallel, but in that case they represent the same plane—in the projective plane, the two lines are not distinct.

Now, in this model of the projective plane, we’re really working with equivalence classes of points in $mathbb R^3setminus {mathbf 0}$: The homogeneous coordinate vectors $(X,Y,Z)$ and $(kX,kY,kZ)$ correspond to the same point in $mathbb P^2$ and it’s the ratios $X:Y:Z$ that serve to distinguish distinct points in $mathbb P^2$. The intersection of the two lines $mathbf l^Tmathbf x=0$ and $mathbf m^Tmathbf x=0$ is, per the above, then, has homogeneous coordinates with ratios $bc'-b'c:ca'-c'a:ab'-a'b$, which are not all zero if the lines are distinct.