Course MT3818 Topics in Geometry

## Isometries of *n*-dimensional space

We are now in a position to calculate the *group of Euclidean geometry*.

**Definition**

If *a* ∈ **R**^{n} then the map *x* ↦ *x* + *a* is called a **translation** and is denoted by *T*_{a}.

Note that translations and orthogonal (linear) transformations are both isometries.

In fact we have the following important result.

**Theorem**

*The group I*(**R**^{n})* of all isometries of ***R**^{n} consists of composites T_{a} ∘ L where T_{a} is a translation and L is an orthogonal map.

**Proof**

If *f* : **R**^{n} **R**^{n} is an isometry then let *f*(**0**) = *a*.

Then *T*_{-a} ∘ *f* maps **0** to **0**. Our result will then follow from the following:

**Lemma**

An isometry *L* which fixes **0** is a linear map.

**Proof of lemma**

Since a straight line can be defined in terms of minimising distance, any isometry maps straight lines to straight lines.

Alos parallel lines are mapped to parallel lines and hence parallelograms are mapped to parallelograms.

Hence *L*(*a* + *b*) = *L*(*a*) + *L*(*b*) since addition is defined using the parallelogram law.

To show that *L*(*λ**a*) = *λ**L*(*a*) observe that this is true for *λ* = 1 and hence (by adding) for any *λ* ∈ **Z** and hence for *λ* ∈ **Q**. Note that any isometry is clearly continuous and so continuity then allows us to extend it to the whole of **R**.

This completes the proof of the lemma and (since any length preserving linear map is orthogonal) of the theorem.

Note that the set of all orthogonal transformations forms a subgroup of *I*(**R**^{n}). We also have:

**Theorem**

*The set of all translations forms a normal subgroup of I*(**R**^{n})* which is isomorphic to the group ***R**^{n} under addition.

**Proof**

Clearly the set of translations forms a subgroup isomorphic to **R**^{n}. To show it is normal, conjugate a translation *T*_{b} by an element *f* = *T*_{a} ∘ *L* of *I*(**R**^{n}).

*f*^{-1}∘ *T*_{b} ∘ *f*(*x*) = *L*^{-1} ∘ *T*_{-a} ∘ *T*_{b}∘ *T*_{a} ∘ *L*(*x*) = *L*^{-1}(*b* + *L*(*x*)) = *L*^{-1}(*b*) + *x* which is translation by *L*^{-1}(*x*) and so is in the subgroup.

**Remark**

In fact the quotient group is isomorphic to the group *O*(*n*). However, in general the subgroup *O*(*n*) of *I*(**R**^{n}) is *not* a normal subgroup and so the group of isometries is *not* the direct product of these two subgroups.

In fact "as a space" *I*(**R**^{n}) is **R**^{n} × *O*(*n*) but the group multiplication is *not* componentwise.

JOC March 2003