Fractional linear and linear
A function of the form
where ad - bc 0 is sometimes called a fractional linear transformation or a bilinear transformation. I usually use the name Mobius transformation.
In what sense are Mobius transformations linear transformations? They're nonlinear functions unless b = c = 0. And yet they're analogous to linear transformations. For starters, the condition ad - bc 0 appears to be saying that a determinant is non-zero, i.e. that a matrix is non-singular.
The transformation g is closely associated with the matrix
but there's more going on than a set of analogies. The reason is that Mobius transformation are linear transformations, but not on the complex numbers .
When you're working with Mobius transformations, you soon want to introduce . Things get complicated if you don't. Once you add theorems become much easier to state, and yet there's a nagging feeling that you may be doing something wrong by informally introducing . This feeling is justified because tossing around without being careful can lead to wrong conclusions.
So how can we rigorously deal with ? We could move from numbers (real or complex) to pairs of numbers, as with fractions. We replace the complex number z with the equivalence class of all pairs of complex numbers whose ratio is z. The advantage of this approach is that you get to add one special number, the equivalence class of all pairs whose second number is 0, i.e. fractions with zero in the denominator. This new number system is called P(), where P" stands for projective."
Mobius transformations are projective linear transformations. They're linear on P(), though not on .
When we multiply the matrix above by the column vector (z 1)T we get
and since our vectors are essentially fractions, the right hand side corresponds to g(z) if the second component of the vector, cz + d, is not zero.
If cz + d = 0, that's OK. Everything is fine while we're working in P(), but we get an element of P() that does not correspond to an element of , i.e. we get .
We've added to the domain and range of our Mobius transformations without any handwaving. We're just doing linear algebra on finite complex numbers.
There's a little bit of fine print. In P() we can't allow both components of a pair to be 0, and non-zero multiples of the same vector are equivalent, so we're not quite doing linear algebra. Strictly speaking a Mobius transformation is a projective linear transformation, not a linear transformation.
It takes a while to warm up to the idea of moving from complex numbers to equivalence classes of pairs of complex numbers. The latter seems unnecessarily complicated. And it often is unnecessary. In practice, you can work in P() by thinking in terms of until you need to have to think about . Then you go back to thinking in terms of P(). You can think of P() as with a safety net for working rigorously with .
Textbooks usually introduce higher dimensional projective spaces before speaking later, if ever, of one-dimensional projective space. (Standard notation would write P^1() rather than P() everywhere above.) But one-dimensional projective space is easier to understand by analogy to fractions, i.e. fractions whose denominator is allowed to be zero, provided the numerator is not also zero.
I first saw projective coordinates as an unmotivated definition. Good morning everyone. We define Pn() to be the set of equivalence classes of n+1 where ...." There had to be some payoff for this added complexity, but we were expected to delay the gratification of knowing what that payoff was. It would have been helpful if someone had said The extra coordinate is there to let us handle points at infinity consistently. These points are not special at all if you present them this way." It's possible someone did say that, but I wasn't ready to hear it at the time.
Related postsThe post Fractional linear and linear first appeared on John D. Cook.