In the lesson on reflections, we defined the charm of an isometry to be the minimal number of mirrors required to express the isometry as the composition of their reflections. We showed that the charm never exceeds 4. Here we apply what we have studied to show that charm never exceeds 3, and identify the most charming isometries, those with charm equal to 3.
The method we shall use is that of "elimination" of all possibilities. This requires some care that we don’t overlook a case. Many a theorem acquired a bad proof, which overlooks an obscure case.
Recall that the identity has charm 0.
If an isometry, $ \alpha = \sigma_\ell $, is a reflections then its charm is just 1.
If an isometry, $ \alpha = \sigma_k \sigma_h $, is the product of two reflection, the two lines $ h, k $ can be in one of three positions
If $ h=k $ then $ \alpha = \sigma^2 = \iota $.
Else if $ Q = (hk) $ then $ \alpha = \rho_{Q,\theta}$, where $ \theta = 2 \angle kh $ .
Else $ h || k $ then $ \alpha = \tau_D $, where $ D = 2 (K-H) $ and $ (KH) $ is perpendicular to the parallel mirrors, with $ K = (k\ell_{KH}), H = (h\ell_{KH}) $ .
Next, suppose $\alpha = \sigma_h \sigma_{g} \sigma_f$. Then the three lines are either different or two are the same. Suppose two of the three are the same:
If $ h=g $ then $ \alpha = \sigma_h^2 \sigma_{g} = \sigma_g $.
Else if $ g = f $ then $ \alpha = \sigma_h $
Else $ h = f $ then $ \alpha = \sigma_{h} \sigma_{g} \sigma_{h}^{-1} = \sigma_{g'} $ where $ g' = \sigma_h(g) $ by the conjugation theorem.
So, in all these cases the charm of $ \alpha $ is 1, and we can assume that all three lines are different. The exactly one of three cases occur, all three are concurrent, all three are parallel (and so concurrent at infinity), or not.
If they are concurrent at $ Q = (fg) =(gh) $ then we can group the three reflections to equal a reflection $ \sigma_h $ following a rotation, or a rotation following a reflection. Let us recalibrate the latter case choosing as mirror $ f $ to replace $ g $. Then there is an "after mirror" so that $ \sigma_{f} \sigma_{g} = \sigma_a \sigma_f $. Substituting we get $ \alpha = (\sigma_a \sigma_f)\sigma_{f} = \sigma_a (\sigma_{f} \sigma_{f}) = \sigma_a $ .
What if we had recalibrated the other way, obtaining $ \alpha = \sigma_b $ for an appropriate "before mirror"?
If we calculate (exercise) the two mirrors exactly, using the recalibration theorem, the we will find that $ a = b $, as expected.
Next, suppose the three mirrors are concurrent at an ideal point, i.e. all three are mutually parallel. Then (exercise) apply the recalibration theorem for translations in an analogous way.
This leaves the case that the three mirrors are not concurrent in either sense and form, a possibly ideal, triangle. These isometries are the most charming ones requires us to settle the case that an isometry is the product of four reflections. To reduce notational clutter we shall use subscripts.
Suppose $ \alpha = \sigma_4 \sigma_3 \sigma_2 \sigma_1 $ with mirrors $ m_4, m_3, m_2, m_1 $ respectively.
First we dispose of the case that two of the mirrors are equal.
If two adjacent mirrors are equal, then the square of an involution reduces the $ \alpha $ to the product of two reflections, treated above.
If either $ m_4 = m_2 $ or $ m_3 = m_1 $ then $ \alpha $ is the product of two reflections, one of which is the conjugate ot $ m_3 $ or $ m_2 $ respectively.
If $ m_4 = m_1 $ then $ \alpha $ is the product of the conjugate of $ m_3 $ and of $ m_2 $ by the same reflection. (Exercise.)
Finally, we may assume that all four mirrors are different. Here the use ideal centers of rotation for translations pays off. Let $ A = (m_4 m_3) , B = (m_2 m_1)$ , whether neither, one or both are ideal. (Exercise, draw pictures of these four cases ). Now either $ A = B $ or not.
Suppose first that the two centers are the same. Note that means that all four lines are either all parallel, or all concurrent. In the parallel case we use the additive algebra of parallel vectors to discover which translation $ \alpha $ collapses to. In the concurrent case, we use the clock arithmetic of rotations about the same center to discover the exact angle for the rotation $ alpha $. (Exercises.)
This leaves the case that $ A \ne B $ and so $ m = (AB) $ is a well defined line. (Exercise: what is $ m $ in each of the four cases that one, both or none of the two points are ideal? )
Now, recalibrate $ \sigma_4 \sigma_3 = \sigma_a \sigma_m $ and $ \sigma_2 \sigma_1 = \sigma_m \sigma_a $ for some appropriate "before" and "after" mirrors. Their product is then another rotation, translation, (Exercise: when is it $ \iota $ ?) i.e. $ \alpha = \sigma_a \sigma_b $.
There is one case that the previous paragraph doesn’t make senses. Suppose both $ A \ne B $ are ideal. The $ m = \infty $ the ideal line. What does it mean to reflect in the ideal line? But if we examine this case separately, all we’re saying is that the first two mirrors are parallel, and the second two mirrors are parallel, but the two sets are not parallel. Since each pair is a translation, and we know how vectors add, we see that the product of four reflections again collapse into a single translation. (Exercise: Why is this translation not the identity.)
What we have now proved is that an isometry is
The identity (charm 0)
A reflection (charm 1)
A rotation or translations (charm 2)
A mystery (charm 3).
Further, we can distinguish between a rotation and a translation by counting the fixed points (1 respectively 0).
In the next and final section we discover the identity of the the mysterious, and most charming isometries.