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As in the last example, we have direct and opposite symmetries.
These are of the form f = Ta
L with L ∈ SO(3).
They can be of three kinds:
Lemma
Let L be a rotation about the line containing the vector b. Then Ta
L is a screw unless a and b are perpendicular.
Proof
Write a = λb + c with c perpendicular to b. Then f = Tλb
(Tc
L).
Now Tc
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.

These are of the form f = Ta
L with L ∈ O(3) - SO(3).
They can be of three kinds:
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
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 = λB + c with c a vector parallel to P.
Then f = Ta
RP = Tc
(Tλb
RP) and Tλb
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
L is also a rotatory reflection.
Proof
Let L = RP
Rotb where the vector b is perpendicular to the plane P. Write a = λb + c with c ∈ P.
Then f = Ta
L = (Tλb
RP)
(Tc
Rotb) since Tc commutes with RP.
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.
| 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 | |
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