Download Isometries of the plane

Isometries of the plane
Mikael Forsberg
August 23, 2011
Abstract
Här följer del av ett dokument om Tesselering som jag skrivit för en
annan kurs. Denna del handlar om isometrier och innehåller bevis för
att en isometri måste vara en sammansättning av en ortogonal avbildning
(ortogonal matris) och en translation. (parallellförflytting). I denna kurs
så behöver man inte gå så här djupt utan detta dokument kan lugnt
hoppas över. Den som strävar efter djupare förståelse kan dock ha viss
glädje av dokumentets innehåll...
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Introduction to isometry
We begin by defining the concept of Isometry:
Definition 1.1. By an Isometry of a subset of the euclidean space Rn we mean
a map I : Rn 7→ Rn which preserves length, that is
||x − y|| = ||I(x) − I(y)||,
for every pair of vectors x and y in Rn .
In spite of its general appearance an isometry is a rather restricted object. In
fact in what follows we prove that an isometry is allways affine with a particular
structure.
Definition 1.2. A map G : Rn 7→ Rn is affine if there is a linear map L and
x0 ∈ Rn such that
G(y) = L(y) + x0 ,
for every y in Rn .
Consider now an isometry F which has the origin as a fix point, that is,
F (0) = 0. Then we have
Lemma 1.3. Suppose F is an isometry with F (0) = 0, then F is linear.
Proof. Taking y = 0 in the definition of an isometry we get that ||F (x)|| = ||x||
for every x ∈ Rn . Now, we have that
||x − y||2 = (x − y) • (x − y) = ||x||2 − 2x • y + ||y||2 ,
from which we get that
||F (x)||2 − 2F (x) • F (y) + ||F (y)||2 = ||x||2 − 2x • y + ||y||2 ,
for every x and y in Rn , and so we have
F (x) • F (y) = x • y.
(1)
This means, of course, that an isometry preserves the scalar product as well as
the length. Now let ej , j = 1, . . . , n be the standard orthonormal basis in Rn .
Then by applying (1) we have that
F (ej ) • F (ek ) = ej • ek = δjk ,
which means that vj = F (ej ) , j = 1, . . . , n is an orthonormal basis in Rn . For
each x, y we have again by (1) That
F (x) • vj = x • ej = xj ,
F (y) • vj = yj ,
F (x + y) • vj = xj + yj ,
and hence F (x + y) and F (x) + F (y) have the same coordinates and so we
conclude that
F (x + y) = F (x) + F (y).
By a similiar argument we easily get that F (cx) = cF (x) and so we have proven
that any isometry which leaves the origin fixed is linear.
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Let us note that by elementary linear algebra we can prove that the condition
(1) is equivalent to F being an orthogonal transformation; if F is expressed as
a matrix AF it will be an orthogonal matrix, and as such it possesses a lot of
nice properties, as is well known from linear algebra. For instance we have that
T
A−1
F = AF from which it follows that
(det AF )2 = det(AF ) det(ATF ) = det(AF ATF ) = det(I) = 1
and so, any orthogonal matrix has determinant either 1 or −1.
Another important consequence of AF being orthogonal is that its column
vectors (which are precisly the vectors vj ) are an orthonormal basis for Rn . And
the converse is also true: a matrix which have orthonormal set of columnvectors
is indeed an orthogonal matrix.
We are now in position to prove our main theorem.
Theorem 1.4. A transformation G is an isometry of Rn if and only if G is an
affine transformation of the form G(x) = L(x) + x0 for every x ∈ Rn where L
is orthogonal.
Proof. Let G be an isometry of Rn . Let F (x) = G(x) − G(0) for every x ∈ Rn .
Then it is easy to see that F is an isometry and evidently we also have F (0) = 0.
By 1.3we therefore have that F is linear and by the discussion above we have
also that F has to be orthogonal.
Conversely, let G(x) = L(x) − x0 , where L be orthogonal. Then L(x) • L(y) =
x • y for every x, y ∈ Rn . If we take x = y we get
||L(x)||2 = L(x) • L(x) = x • x = ||x||2 ,
hence, ||L(x)|| = ||x||. Replacing x by x − y we have that
||L(x) − L(y)|| = ||L(x − y)|| = ||x − y||,
for every x, y ∈ Rn and so we have that L is an isometry. Since ||G(x)−G(y)|| =
||L(x) − L(y)|| we conclude that G is an isometry, which finishes the proof.
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Examples
Let us now consider the special case where n = 2.
Exempel 2.1. If L is equal to the identity matrix I and x0 = 0 then we have
the identity isometry. A matrix representation is
x
1 0
x
I
=
y
0 1
y
If x0 6= 0 then G is a pure translation and in matrix notation we get
x
1 0
x
x0
G
=
+
.
y
0 1
y
y0
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Exempel 2.2. If L is an orthogonal matrix with positive determinant (= +1)
then G is a pure rotation if x0 = 0. In matrix notation we have for a rotation
by an angle t:
x
cos t sin t
x
G
=
.
y
− sin t cos t
y
Isometries of the above types are called direct or even isometries.
Övning 2.1. Show that a nonpure rotation, i.e. a rotation followed by a translation, is a pure rotation.
The following two isometries are called indirect or odd isometries.
Exempel 2.3. If G = R where R is an orthogonal matrix with determinant
equal to −1 then G is called a pure reflection. In matrix notation a reflection
becomes
x
cos 2t
sin 2t
x
G
=
,
y
sin 2t − cos 2t
y
where t is the angle between the x-axis and the line of reflection.
Övning 2.2. Prove the matrix representation of a reflection.
Exempel 2.4. If G is a translation followed by a reflection in the line of translation then the result is called a glide reflection. If we express this in matrix
form we get:
x
cos 2t0
sin 2t0
x
x0
G
=
+
,
y
sin 2t0 − cos 2t0
y
y0
where t0 = arctan y0 /x0 .
Övning 2.3. Show that a translation followed by a reflection in a line not
necessarily parallell to the line of tranlation is in fact a glide reflection.
Proposition 2.5. An isometry I has a fix point iff I is a rotation or a reflection.
Proof. If I is a reflection or a rotation we know that it has no translation and
therefore is an orthogonal matrix. Any matrix is linear and so has 0 as fix point.
Assume now that I has a fixpoint;
I(x) := L(x) + x0 = x,
for some orthogonal mapping L. If x0 = 0 we have immediately that I is either
a rotation or a reflection. Assume now that x0 6= 0. For L this means that
L(x) = x − x0 ,
from which it follows that L is not linear:
L(a + b) = (a + b) − x0 6= L(a) + L(b) = a − x0 + b − x0 = a + b − 2x0 .
This is a contradiction to the fact that L was orthogonal! The only possibility
to have an isometry with a fixpoint is therefore that it has no translational
component and so is a pure orthogonal mapping
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Consider again Theorem 1.4. With the notation of the examples it is not
difficult to see that the theorem states that any isometry is one of the the above
four types.
Composition of two or more isometries will of course result in an isometry.
In fact isometries of the plane together with composition will constitute a group,
an object which we will study in the next chapter.
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