[FOM] A new geometry (in Klein's sense)
dana.scott at cs.cmu.edu
Sat Feb 18 01:02:45 EST 2006
Giovanni Lagnese replied on February 17, 2006 7:16:26 AM PST:
> Dana Scott wrote:
>> I think it is easy to prove that when M = R^n and the distance
>> is the usual n-dimensional Euclidean metric, then the group
>> of such mappings is just the usual group rigid motions of
>> Euclidean space.
> f(v)=2v is in the group of such mappings but it is not a rigid motion.
That is very true, and I should have known better. Rigid motions
of Euclidean space preserve distance; Euclidean motions preserve
equality of distances. Similitudes (= change of scale) do not
preserve distances, but they do preserve equality of distances.
We want to show that Euclidean motions are compositions of rigid
motions and similitudes.
In 2-D, one notes first that Euclidean motions must preserve
the perpendicular bisector of a segment, because it is the locus
of points equidistant from the end points. Hence, euclidean motions
preserve lines. Also, such transformations must preserve pairs
of perpendicular lines.
So, starting with such a motion, one can compose it with a translation
and a rotation so that a pair of perpendicular lines is left fixed
(but not necessarily pointwise fixed). Think of these as a pair of
coordinate axes intersecting at an "origin" called "O". The point
O is left fixed by the altered motion.
Designate one point on one axis not at the origin as a "unit" point.
If it transforms to the other side of the origin, then compose the
motion with reflection about the perpendicular axis, so the image of
the point is on the same side of the origin.
Next, choose a unit point on the other axis at the same distance from
the origin as the first chosen unit point on the first axis. Again,
compose with a reflection so under the motion it stays on the same side
of the origin.
Call the two unit points "X" and "Y". By construction we have
d(X,O) = d(Y,O). Under the (altered) motion (altered after composition
with those rigid motions) we have d(f(X),O) = d(f(Y),O). The positive
alpha = d(f(X),O)/d(X,O)
should be the uniform factor of proportionality for the transformation.
But by the preservation of equality of distances, we also have:
alpha = d(f(Y),O)/d(Y,O).
So, compose f with the similitude about the origin O with factor 1/
The altered transformation leaves the three points O, X, and Y each
If we can show that a Euclidean motion leaving each of three non-
collinear points fixed is the identity, then our original motion is
indeed the composition of a rigid motion with a similitude. What we
need is a:
LEMMA. Three circles with non-collinear centers can intersect in
at most one point.
Proof: If two circles intersect in two distinct points, then the
line joining their centers is the perpendicular bisector of the
segment between them. Hence, if another circle passes through
these same two points, its center is also on the line joining the
centers of the first two circles. Q.E.D.
Note that a Euclidean motion has to preserve circles. (Why?) If the
center of the circle is left fixed, then the circle is mapped onto
itself. Hence, if P is any point in the plane, then the circle through
P with center O has to be mapped onto itself. Similarly for the circles
with centers X and Y also passing through P. But f(P) has to be on each
of these three circles. THEREFORE, P = f(P). Q.E.D.
Note that I have used the Euclidean geometry of R^2 over and over here.
The argument can be adapted to R^n as well.
But what of other metric spaces and their equality-of-distance
preserving mappings? Do we get interesting groups for them?
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