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Every finite subgroup of isometries of n-dimensional space

  1. Oct 26, 2007 #1
    ... fixes at least one point.

    I recently came upon the proof in a book and I didn't quite understand the notion of "rigid motion", and I was wondering if you could help clarify it for me. Is it just "the vertices must stay in the given order", as used in symmetries of polygons? I've attached the proof.

    Thanks.
     

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    Last edited: Oct 26, 2007
  2. jcsd
  3. Oct 26, 2007 #2
    A "rigid motion" is also called an isometry. An isometry is a function [tex]f:\mathbb{R}^n \mapsto \mathbb{R}^n[/tex] so that [tex]|\bold{x}-\bold{y}| = |f(\bold{x}) - f(\bold{y})|[/tex]. Where [tex] | \ |[/tex] is the Euclidean metric. All this means it is a function such that the distance between any two points remains the same after the function is performed. Thus, in a sense it is "rigid", i.e. it leaves the object in tact. So reflections, rotations, translations are simple examples of functions which leave objects unchanged.

    Given a set (or an object) in space a "symettry" is a rigid motion which leaves the object unchanged, i.e. it is its own image under the map. In that case we can define a "symettery group" on an object as the set of all symettries on the object.

    I think what you are trying to prove is that the group of symettries on a finite set is either the dihedral group or a cyclic group.
     
  4. Oct 26, 2007 #3
    Alright, thank you. That makes sense. Basically, a rigid motion will preserve distances and angles (in terms of vectors).

    Can you clarify what it means when the Orthogonal group, [itex]O(n)[/itex], fixes the origin?
     
    Last edited: Oct 27, 2007
  5. Oct 27, 2007 #4

    morphism

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    It means every isometry in O(n) leaves 0 fixed. The elements of O(n) aren't 'vectors' - they're maps/matrices.
     
  6. Oct 27, 2007 #5
    What they are saying in the book is this:

    Let G be your finite group of isometries/rigid motions on a vector space V. Let v ∈ V be arbitrary. Define

    [tex]w := \frac{1}{G} \sum_{g \in G} g(v)[/tex]

    now for any h ∈ G, we have h(w)=w:

    [tex]h(w) = h\left( \frac{1}{G} \sum_{g \in G} g(v) \right) = \frac{1}{G} \sum_{g \in G} hg(v) = w[/tex]

    The second equality holds because (I think) an isometry of a Hilbert space is automatically an affine map (it maps affine combinations to affine combinations, and our sum there is an affine combination since the coefficients sum to 1).

    The last equality holds because multiplication with a group element is bijective, so it is the same sum, just reordered.
     
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