Course MT3818 Topics in Geometry

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Isometries in 3 dimensions

As in the last example, we have direct and opposite symmetries.

  1. Direct symmetries

    These are of the form f = Tacomp L with L belongs SO(3).
    They can be of three kinds:

    1. A rotation (about any line in R3),
    2. A translation,
    3. A screw translation (or screw): a rotation about some line followed by a translation parallel to that line.

    Proof
    If L is the identity, then f is a translation, otherwise the following result handles things.

    Lemma
    Let L be a rotation about the line containing the vector b. Then Tacomp L is a screw unless a and b are perpendicular.

    Proof
    Write a = lambdab + C with C perpendicular to b. Then f = Tlambdab comp (TC comp L).
    Now TC comp L is rotation about some axis parallel to b since we are reduced to the two-dimensional situation in the plane perpendicular to b.
    Hence f is a screw translation unless lambda = 0 in which case it is a rotation.


  2. Opposite symmetries

    These are of the form f = Tacomp L with L belongs O(3) - SO(3).
    They can be of three kinds:

    1. A reflection (about any plane in R3),
    2. A glide reflection: reflection in a plane P followed by a translation parallel to P,
    3. A rotatory reflection: reflection in a plane P followed by a rotation about an axis perpendicular to P.

    Proof
    An element of O(3) - SO(3) is either reflection a plane or a rotatory reflection. The first of these two possibilities is handled by:

    Lemma
    Let RP be reflection in the plane P. Then Ta comp RP is reflection in a plane if a is perpendicular to P and is a glide reflection otherwise.

    Proof
    Let b be a vector perpendicular to P. Write a = lambdaB + C with C a vector parallel to P.
    Then f = Ta comp RP = TC comp (Tlambdabcomp RP) and Tlambdabcomp RP is reflection in a plane parallel to P since this is essentially the two-dimensional situation considered ealier.
    So if the vector C = 0 then f is a reflection and otherwise it is a glide reflection.


    The second possibility is handled by:

    Lemma
    Let L be a rotatory reflection. Then Ta comp L is also a rotatory reflection.

    Proof
    Let L = RP comp Rotb where the vector b is perpendicular to the plane P. Write a = lambdab + C with C belongs P.
    Then f = Ta comp L = (Tlambdab comp RP) comp (TC comp Rotb) since TC commutes with RP.
    Since lambdab is perpendicular to P, the first bracket is reflection in a plane parallel to P.
    Since C is perpendicular to b, the second bracket is rotation about an axis parallel to b.
    Hence this is a rotatory reflection as required.

    This completes the classification of the opposite symmetries.

Remarks

  1. We get the following summary about isometries of R3.
    Fixed pointDirect symmetryOpposite symmetry
    NoneTranslation or ScrewGlide reflection
    dim 0 Rotatory reflection
    dim 1Rotation 
    dim 2 Reflection
    dim 3Identity 

  2. One can also classify the different kinds of symmetry by looking at how they can be written as products of reflections in planes. Note that a product of an even number of reflections is a direct symmetry, a product of an odd number is opposite.
    1. Reflection in a plane

    2. A products of two reflections is a translation if the planes are parallel and a rotation (about the line where the planes meet) otherwise.

    3. A product of three reflections is a rotatory reflection if the three planes meet in a point. If two of the planes are parallel and meet the third or if the three planes are parallel, then we get a glide reflection.

    4. To get a screw translation we need a product of four reflections.

  3. In fact any symmetry of Rn can be written as a product of at most n + 1 reflections in (n - 1)-dimensional hyperplanes.


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JOC February 2003