1Euclidean geometry
IB Geometry
1.1 Isometries of the Euclidean plane
The purpose of this section is to study maps on R
n
that preserve distances, i.e.
isometries of
R
n
. Before we begin, we define the notion of distance on
R
n
in the
usual way.
Definition ((Standard) inner product). The (standard) inner product on
R
n
is
defined by
(x, y) = x · y =
n
X
i=1
x
i
y
i
.
Definition (Euclidean Norm). The Euclidean norm of x ∈ R
n
is
kxk =
p
(x, x).
This defines a metric on R
n
by
d(x, y) = kx − yk.
Note that the inner product and the norm both depend on our choice of
origin, but the distance does not. In general, we don’t like having a choice of
origin — choosing the origin is just to provide a (very) convenient way labelling
points. The origin should not be a special point (in theory). In fancy language,
we say we view R
n
as an affine space instead of a vector space.
Definition (Isometry). A map f : R
n
→ R
n
is an isometry of R
n
if
d(f(x), f(y)) = d(x, y)
for all x, y ∈ R
n
.
Note that
f
is not required to be linear. This is since we are viewing
R
n
as an affine space, and linearity only makes sense if we have a specified point
as the origin. Nevertheless, we will still view the linear isometries as “special”
isometries, since they are more convenient to work with, despite not being special
fundamentally.
Our current objective is to classify all isometries of
R
n
. We start with the
linear isometries. Recall the following definition:
Definition (Orthogonal matrix). An
n ×n
matrix
A
is orthogonal if
AA
T
=
A
T
A = I. The group of all orthogonal matrices is the orthogonal group O(n).
In general, for any matrix A and x, y ∈ R
n
, we get
(Ax, Ay) = (Ax)
T
(Ay) = x
T
A
T
Ay = (x, A
T
Ay).
So A is orthogonal if and only if (Ax, Ay) = (x, y) for all x, y ∈ R
n
.
Recall also that the inner product can be expressed in terms of the norm by
(x, y) =
1
2
(kx + yk
2
− kxk
2
− kyk
2
).
So if
A
preserves norm, then it preserves the inner product, and the converse
is obviously true. So
A
is orthogonal if and only if
kA
x
k
=
k
x
k
for all x
∈ R
n
.
Hence matrices are orthogonal if and only if they are isometries.
More generally, let
f(x) = Ax + b.
Then
d(f(x), f(y)) = kA(x − y)k.
So any
f
of this form is an isometry if and only if
A
is orthogonal. This is not
too surprising. What might not be expected is that all isometries are of this
form.
Theorem. Every isometry of f : R
n
→ R
n
is of the form
f(x) = Ax + b.
for A orthogonal and b ∈ R
n
.
Proof. Let f be an isometry. Let e
1
, ··· , e
n
be the standard basis of R
n
. Let
b = f(0), a
i
= f(e
i
) − b.
The idea is to construct our matrix
A
out of these a
i
. For
A
to be orthogonal,
{a
i
} must be an orthonormal basis.
Indeed, we can compute
ka
i
k = kf(e
i
) − f(0)k = d(f(e
i
), f(0)) = d(e
i
, 0) = ke
i
k = 1.
For i 6= j, we have
(a
i
, a
j
) = −(a
i
, −a
j
)
= −
1
2
(ka
i
− a
j
k
2
− ka
i
k
2
− ka
j
k
2
)
= −
1
2
(kf(e
i
) − f(e
j
)k
2
− 2)
= −
1
2
(ke
i
− e
j
k
2
− 2)
= 0
So a
i
and a
j
are orthogonal. In other words,
{
a
i
}
forms an orthonormal set. It
is an easy result that any orthogonal set must be linearly independent. Since we
have found n orthonormal vectors, they form an orthonormal basis.
Hence, the matrix
A
with columns given by the column vectors a
i
is an
orthogonal matrix. We define a new isometry
g(x) = Ax + b.
We want to show
f
=
g
. By construction, we know
g
(x) =
f
(x) is true for
x = 0, e
1
, ··· , e
n
.
We observe that g is invertible. In particular,
g
−1
(x) = A
−1
(x − b) = A
T
x − A
T
b.
Moreover, it is an isometry, since
A
T
is orthogonal (or we can appeal to the
more general fact that inverses of isometries are isometries).
We define
h = g
−1
◦ f.
Since it is a composition of isometries, it is also an isometry. Moreover, it fixes
x = 0, e
1
, ··· , e
n
.
It currently suffices to prove that h is the identity.
Let x ∈ R
n
, and expand it in the basis as
x =
n
X
i=1
x
i
e
i
.
Let
y = h(x) =
n
X
i=1
y
i
e
i
.
We can compute
d(x, e
i
)
2
= (x − e
i
, x − e
i
) = kxk
2
+ 1 − 2x
i
d(x, 0)
2
= kxk
2
.
Similarly, we have
d(y, e
i
)
2
= (y − e
i
, y − e
i
) = kyk
2
+ 1 − 2y
i
d(y, 0)
2
= kyk
2
.
Since
h
is an isometry and fixes 0
,
e
1
, ··· ,
e
n
, and by definition
h
(x) = y, we
must have
d(x, 0) = d(y, 0), d(x, e
i
) = d(y, e
i
).
The first equality gives
k
x
k
2
=
k
y
k
2
, and the others then imply
x
i
=
y
i
for all
i
.
In other words, x = y = h(x). So h is the identity.
We now collect all our isometries into a group, for convenience.
Definition (Isometry group). The isometry group
Isom
(
R
n
) is the group of all
isometries of R
n
, which is a group by composition.
Example (Reflections in an affine hyperplane). Let
H ⊆ R
n
be an affine
hyperplane given by
H = {x ∈ R
n
: u · x = c},
where
k
u
k
= 1 and
c ∈ R
. This is just a natural generalization of a 2-dimensional
plane in
R
3
. Note that unlike a vector subspace, it does not have to contain the
origin (since the origin is not a special point).
Reflection in H, written R
H
, is the map
R
H
: R
n
→ R
n
x 7→ x − 2(x · u − c)u
It is an exercise in the example sheet to show that this is indeed an isometry.
We now check this is indeed what we think a reflection should be. Note that
every point in
R
n
can be written as a +
t
u, where a
∈ H
. Then the reflection
should send this point to a − tu.
0
a
a − tu
H
a + tu
This is a routine check:
R
H
(a + tu) = (a + tu) − 2tu = a − tu.
In particular, we know R
H
fixes exactly the points of H.
The converse is also true — any isometry
S ∈ Isom
(
R
n
) that fixes the points
in some affine hyperplane H is either the identity or R
H
.
To show this, we first want to translate the plane such that it becomes a
vector subspace. Then we can use our linear algebra magic. For any a
∈ R
n
, we
can define the translation by a as
T
a
(x) = x + a.
This is clearly an isometry.
We pick an arbitrary a
∈ H
, and let
R
=
T
−a
ST
a
∈ Isom
(
R
n
). Then
R
fixes
exactly
H
0
=
T
−a
H
. Since 0
∈ H
0
,
H
0
is a vector subspace. In particular, if
H = {x : x · u = c}, then by putting c = a · u, we find
H
0
= {x : x · u = 0}.
To understand
R
, we already know it fixes everything in
H
0
. So we want to see
what it does to u. Note that since
R
is an isometry and fixes the origin, it is in
fact an orthogonal map. Hence for any x ∈ H
0
, we get
(Ru, x) = (Ru, Rx) = (u, x) = 0.
So
R
u is also perpendicular to
H
0
. Hence
R
u =
λ
u for some
λ
. Since
R
is an
isometry, we have
kR
u
k
2
= 1. Hence
|λ|
2
= 1, and thus
λ
=
±
1. So either
λ
= 1, and
R
=
id
; or
λ
=
−
1, and
R
=
R
H
0
, as we already know for orthogonal
matrices.
It thus follow that S = id
R
n
, or S is the reflection in H.
Thus we find that each reflection R
H
is the (unique) isometry fixing H but
not id
R
n
.
It is an exercise in the example sheet to show that every isometry of
R
n
is a composition of at most
n
+ 1 reflections. If the isometry fixes 0, then
n
reflections will suffice.
Consider the subgroup of
Isom
(
R
n
) that fixes 0. By our general expression
for the general isometry, we know this is the set
{f
(x) =
A
x :
AA
T
=
I}
∼
=
O(
n
),
the orthogonal group.
For each
A ∈
O(
n
), we must have
det
(
A
)
2
= 1. So
det A
=
±
1. We use this
to define a further subgroup, the special orthogonal group.
Definition (Special orthogonal group). The special orthogonal group is the
group
SO(n) = {A ∈ O(n) : det A = 1}.
We can look at these explicitly for low dimensions.
Example. Consider
A =
a b
c d
∈ O(2)
Orthogonality then requires
a
2
+ c
2
= b
2
+ d
2
= 1, ab + cd = 0.
Now we pick 0 ≤ θ, ϕ ≤ 2π such that
a = cos θ b = −sin ϕ
c = sin θ d = cos ϕ.
Then
ab
+
cd
= 0 gives
tan θ
=
tan ϕ
(if
cos θ
and
cos ϕ
are zero, we formally say
these are both infinity). So either θ = ϕ or θ = ϕ ±π. Thus we have
A =
cos θ −sin θ
sin θ cos θ
or A =
cos θ sin θ
sin θ −cos θ
respectively. In the first case, this is a rotation through
θ
about the origin. This
is determinant 1, and hence A ∈ SO(2).
In the second case, this is a reflection in the line
`
at angle
θ
2
to the
x
-axis.
Then det A = −1 and A 6∈ SO(2).
So in two dimensions, the orthogonal matrices are either reflections or rota-
tions — those in SO(2) are rotations, and the others are reflections.
Before we can move on to three dimensions, we need to have the notion of
orientation. We might intuitively know what an orientation is, but it is rather
difficult to define orientation formally. The best we can do is to tell whether
two given bases of a vector space have “the same orientation”. Thus, it would
make sense to define an orientation as an equivalence class of bases of “the same
orientation”. We formally define it as follows:
Definition (Orientation). An orientation of a vector space is an equivalence
class of bases — let v
1
, ··· ,
v
n
and v
0
1
, ··· ,
v
0
n
be two bases and
A
be the change
of basis matrix. We say the two bases are equivalent iff
det A >
0. This is an
equivalence relation on the bases, and the equivalence classes are the orientations.
Definition (Orientation-preserving isometry). An isometry
f
(x) =
A
x + b
is orientation-preserving if
det A
= 1. Otherwise, if
det A
=
−
1, we say it is
orientation-reversing.
Example. We now want to look at O(3). First focus on the case where
A ∈ SO(3), i.e. det A = 1. Then we can compute
det(A − I) = det(A
T
− I) = det(A) det(A
T
− I) = det(I − A) = −det(A − I).
So
det
(
A − I
) = 0, i.e. +1 is an eigenvalue in
R
. So there is some v
1
∈ R
3
such
that Av
1
= v
1
.
We set W = hv
1
i
⊥
. Let w ∈ W . Then we can compute
(Aw, v
1
) = (Aw, Av
1
) = (w, v
1
) = 0.
So
A
w
∈ W
. In other words,
W
is fixed by
A
, and
A|
W
:
W → W
is well-defined.
Moreover, it is still orthogonal and has determinant 1. So it is a rotation of the
two-dimensional vector space W .
We choose
{
v
2
,
v
3
}
an orthonormal basis of
W
. Then under the bases
{v
1
, v
2
, v
3
}, A is represented by
A =
1 0 0
0 cos θ −sin θ
0 sin θ cos θ
.
This is the most general orientation-preserving isometry of
R
3
that fixes the
origin.
How about the orientation-reversing ones? Suppose
det A
=
−
1. Then
det(−A) = 1. So in some orthonormal basis, we can express A as
−A =
1 0 0
0 cos θ −sin θ
0 sin θ cos θ
.
So A takes the form
A =
−1 0 0
0 cos ϕ −sin ϕ
0 sin ϕ cos ϕ
,
where
ϕ
=
θ
+
π
. This is a rotated reflection, i.e. we first do a reflection, then
rotation. In the special case where ϕ = 0, this is a pure reflection.
That’s all we’re going to talk about isometries.