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Any invertible map *T* from a vector space *V* to itself leads to a bijection of the associated projective space *P*(*V*).

(Just map [** u**] [

However, maps of the form

**Definition**

If *F* is any field, the quotient group *GL*(*n*, *F*)/{*I* | *F* - {0} } is called the **projective group** and is written *PGL*(*n*, *F*).

The elements of this group are called **projective transformations** or **projectivities**. (The name *projection* is used for something different.)

We will now look at some examples

The easiest (and most important) case is *PGL*(2, **R**).

This is the group which acts on *P*(**R**^{2}) = **R***P*^{1}: the *real projective line*.

In Klein's formulation it is the group of 1-dimensional real projective geometry.

Let represent an element of *PGL*(2, **R**). Note that etc. will represent the same element.

This acts on an ordinary point with homogeneous coordinates [*x*, 1] to give [*ax* + *b* , *cx* + *d*] or equivalently [(*ax* + *b*)/(*cx* + *d*)] provided that *cx* + *d* 0.

That is, the matrix corresponds to the rational map *x* (*ax* + *b*)/(*cx* + *d*) where we interpret this as if *cx* + *d* = 0. Note that we must have ad - bc 0 otherwise the map would not be invertible.

Looking at what happens to the point at infinity = [1, 0], this maps to ^{a}/_{c} (just as we might expect).

**Remarks**

- One may prove that the composite of such rational maps corresponds to the multiplication of matrices.

(Believe it or not this was the way that the English mathematician Arthur Cayley (1821 to 1895) first defined the multiplication of matrices). - Such projective transformations do correspond to projection from a point as seen earlier. (See Exercises 8 Question 3)

The following trick makes studying projectivities easier.

**Definition**

The **standard reference points** on **R***P*^{1} are , 0 ,1 (that is [0, 1], [0, 1] and [1, 1]).

**Theorem**

*There is a unique projective transformation taking any three distinct points to* , 0 ,1.

**Proof**

The map is *x* (*x* - *b*)/(*x* - *a*) . (*c* - *a*)/(*c* - *b*).

**Corollary**

*There is a unique projective transformation taking any three distinct points to any other three distinct points*.

**Proof**

If the map maps *a*, *b*, *c* to , 0 ,1 and maps *a*', *b*', *c*' to , 0 ,1 then the map ^{-1} does what is needed.

Projective maps do not preserve lengths or ratios or even intermediacy. However they do preserve something called *cross-ratio*.

**Definition**

Let *a*, *b*, *c*, *d* be four points of **R***P*^{1}. Let be the map taking *a*, *b*, *c* to , 0, 1. Then the **cross-ratio** (*a* , *b* ; *c* , *d*) is (*d*).

From the above this is (*a* , *b* ; *c* , *d*) = (*d* - *b*)/(*d* - *a*) . (*c* - *a*)/(*c* - *b*).

**Remarks**

- Drawing the points on a line:

the cross ratio is the ratio in which*C*divides*AB*divided by the ratio in which*D*divides*AB*: (^{AC}/_{CB})/(^{AD}/_{DB}). - If
*D*is at infinity this is the ratio*AC*/*BC*( = -*AC*/*CB*). - If the cross-ratio is -1, the four points are said to be
*harmonic*.

If*D*is at infinity then*AC*=*CB*for a harmonic range (as it is called).

**Proof**

If *f* is a projective map and takes *a*, *b*, *c* to , 0, 1 then *f* ^{-1} takes *f*(*a*), *f*(*b*), *f*(*c*) to , 0, 1. The cross ratio (*a* , *b* ; *c* , *d*) is (*d*) and this is the same as *f* ^{-1}(*f*(*d*)) and the result follows.

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