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
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