Isometries in 3 dimensions

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

1. Direct symmetries

   These are of the form f = T_a∘ L with L ∈ 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 T_a∘ L is a screw unless a and b are perpendicular.

   Proof
   Write a = λb + C with C perpendicular to b. Then f = T_λb ∘ (T_C ∘ L).
   Now T_C ∘ 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 λ = 0 in which case it is a rotation.
   
   
   

2. Opposite symmetries

   These are of the form f = T_a∘ L with L ∈ 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 R_P be reflection in the plane P. Then T_a ∘ R_P 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 = λB + C with C a vector parallel to P.
   Then f = T_a ∘ R_P = T_C ∘ (T_λb∘ R_P) and T_λb∘ R_P 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 T_a ∘ L is also a rotatory reflection.

   Proof
   Let L = R_P ∘ Rot_b where the vector b is perpendicular to the plane P. Write a = λb + C with C ∈ P.
   Then f = T_a ∘ L = (T_λb ∘ R_P) ∘ (T_C ∘ Rot_b) since T_C commutes with R_P.
   Since λb 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.
   

1. We get the following summary about isometries of R³.
   

   Fixed point	Direct symmetry	Opposite symmetry
   None	Translation or Screw	Glide reflection
   dim 0		Rotatory reflection
   dim 1	Rotation
   dim 2		Reflection
   dim 3	Identity

   
   
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.