In $ \scriptstyle D_n $, explain geometrically why a reflection followed by a reflection must be a rotation.
When two reflections are performed with respect to a certain axis, an invariant point with respect to the point(s) being reflected is formed in the intersection of the lines of reflection. An angle is formed as well, at this intersection. It turns out that the double of this angle is the very angle through which an equivalent rotation is performed. We may choose any intersecting orientation for the two lines, that is to say, any angle between the lines of reflection is possible. In the case of the two lines of reflection being equal, the identity transformation is performed, and is congruent to a 0 radian rotation.
Thinking about it visually, imagine you have a 2-D shape with a red side and a green side, red side facing up. If only a rotation was performed on the shape, the red side should still be facing up. This is why two reflections are needed: to preserve the cyclic order of the vertices of the shape. If cyclic order doesn't matter (for example, if we are only concerned with the placement of the edges and not the specific vertices of a polygon), then a reflection in a line of symmetry is congruent to a rotation. The dihedral groups of order $ \scriptstyle2n $ are somewhat special in the fact that they each have $ \scriptstyle n $ unique bilateral lines of symmetry.
Mathematically, it's easy to prove that the composition of two reflections is a rotation. With respect to the positive Cartesian x-axis, a reflection of a point in a line at an angle $ \scriptstyle\theta $ can be expressed by a matrix transformation of the form:
$ \begin{bmatrix} \scriptstyle cos(2\theta) & \scriptstyle sin(2\theta) \\ \scriptstyle sin(2\theta) & \scriptstyle-cos(2\theta) \end{bmatrix} $.
And a rotation about the origin through an angle $ \scriptstyle\theta $ (measuring anticlockwise from the positive x-axis) may be represented as:
$ \begin{bmatrix} \scriptstyle cos(\theta) & \scriptstyle-sin(\theta) \\ \scriptstyle sin(\theta) & \scriptstyle cos(\theta) \end{bmatrix} $.
Certainly, when we compose two reflections, we obtain
$ \begin{bmatrix} \scriptstyle cos(2\theta) & \scriptstyle sin(2\theta) \\ \scriptstyle sin(2\theta) & \scriptstyle-cos(2\theta) \end{bmatrix} \begin{bmatrix} \scriptstyle cos(2\phi) & \scriptstyle sin(2\phi) \\ \scriptstyle sin(2\phi) & \scriptstyle-cos(2\phi) \end{bmatrix} $
$ =\begin{bmatrix} \scriptstyle cos(2\theta)cos(2\phi)+sin(2\theta)sin(2\phi) & \scriptstyle cos(2\theta)sin(2\phi)-sin(2\theta)cos(2\phi) \\ \scriptstyle sin(2\theta)cos(2\phi)-cos(2\theta)sin(2\phi) & \scriptstyle sin(2\theta)sin(2\phi)+cos(2\theta)cos(2\phi) \end{bmatrix} =\begin{bmatrix} \scriptstyle cos(2(\theta-\phi)) & \scriptstyle sin(2(\phi-\theta)) \\ \scriptstyle sin(2(\theta-\phi)) & \scriptstyle cos(2(\theta-\phi)) \end{bmatrix} $
$ =\begin{bmatrix} \scriptstyle cos(2(\theta-\phi)) & \scriptstyle-sin(2(\theta-\phi)) \\ \scriptstyle sin(2(\theta-\phi)) & \scriptstyle cos(2(\theta-\phi)) \end{bmatrix} $
which represents a rotation about the origin through the angle $ \scriptstyle2(\theta-\phi) $.
- --Nick Rupley 03:10, 25 January 2009 (UTC)
Nick, thanks so much for the explanation, that makes so much more sense!
- --Emily Raymond 11:03, 20 January 2009 (UTC)
Agreed. I was not sure how to do this one either and this explanation helped a lot! --Johns121 16:30, 29 January 2009 (UTC)
