Course MT3818 Topics in Geometry

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Isometries of n-dimensional space

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

Definition

If a belongs Rn then the map x goesto x + a is called a translation and is denoted by Ta.

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

In fact we have the following important result.

Theorem
The group I(Rn) of all isometries of Rn consists of composites Ta comp L where Ta is a translation and L is an orthogonal map.

Proof
If f : Rn rarrow Rn is an isometry then let f(0) = a.
Then T-a comp 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(lambdaa) = lambdaL(a) observe that this is true for lambda = 1 and hence (by adding) for any lambda belongs Z and hence for lambda belongs 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(Rn). We also have:

Theorem
The set of all translations forms a normal subgroup of I(Rn) which is isomorphic to the group Rn under addition.

Proof
Clearly the set of translations forms a subgroup isomorphic to Rn. To show it is normal, conjugate a translation Tb by an element f = Ta comp L of I(Rn).
f-1comp Tb comp f(x) = L-1 comp T-a comp Tbcomp Ta comp 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(Rn) 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(Rn) is Rn cross O(n) but the group multiplication is not componentwise.


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JOC March 2003