36 1. Various Ways of Representing Surfaces and Examples

a

Ia

b

Ib

c = Ic

a

Ia b

Ib

Figure 1.22. An orientation preserving isometry with no

fixed points is a translation.

Rotations are entirely determined by the centre of rotation and

the angle of rotation, and so are specified by three parameters.

Case 2 : An orientation preserving isometry I with no fixed points

is a translation. The easiest way to see that is to use the complex

algebraic description. Writing Iz = az + b with |a| = 1, we observe

that if a = 1, we can solve az + b = z to find a fixed point for I.

Since no such point exists, we have a = 1, hence I : z → z + b is a

translation.

One can also make a purely synthetic argument for this case; we

show that the intervals [a, Ia] and [b, Ib] must be parallel and of equal

length for every a, b. Indeed, if they fail to be parallel for some a, b,

then their perpendicular bisectors intersect in some point c, as shown

in Figure 1.22. Since [a, Ia, c] and [b, Ib, c] are isosceles triangles, we

have d(a, c) = d(Ia, c) and d(b, c) = d(Ib, c), hence Ic = c since I

preserves orientations.

But I has no fixed point, and so [a, Ia] and [b, Ib] must be parallel;

since I is an isometry, d(Ia, Ib) = d(a, b), and hence the quadrilateral

[a, Ia, Ib, b] is a parallelogram. It follows that the intervals [a, Ia] are

all parallel and of equal length, and so I is a translation.

We only require two parameters to specify a translation; since

the space of translations is two-dimensional, almost every orientation

preserving isometry is a rotation, and hence has a fixed point.

Case 3 : An orientation reversing isometry which possesses a fixed

point is a reflection. Say Ix = x, and fix y = x. Let be the line

bisecting the angle formed by the points y, x, Iy. Using the same