1. Background

In this lesson we define the ultimate building block, the atoms among the isometries. These are the reflections in lines, and unlike the transformations we have studied so far, they do not have a simple formula in vector algebra, like dilatations, translations, and central reflections.

The concept is simple enough. To reflect a point $ X $ across a line $ \ell $, drop the perpendicular from $ X $ to $ \ell $ and extend it an equal distance to the other side of the line. This point transformation is aptly called a reflection, because if $ \ell $ is considered to be a mirror, then the reflected object would look as if it were on the "other side" of the mirror. A point on the mirror line is moved to itself, i.e. it is a fixed point of the reflection. In fact, we will soon see that all the fixed points of a reflection $ \sigma_\ell $ lie on the mirror line $ \ell $ of the reflection.

A reflection does not have an easy, explicit formula that defines it, and so is a new kind of transformation. A graphics computer requires a formula, however, and we next develop one, which might more aptly be called a recipe for a formula, which a computer would be just as happy to use as a single equation.

We begin by emulating Euclid and KSEG with two points, $ A $ and $ B $, that determine the mirror line $ \ell = \ell_{AB} $. This choice is obviously not unique. If we make no further restrictions on which points, and for what lines, then the explicit formula is fierce indeed:

$ \sigma_{\ell_{AB}}(X) = 2A-X+2\frac{(X-A)(B-A)}{(B-A)(B-A)}(B-A) $


where the scaling of the displacement vector $ (B-A) $ is the ratio of two dot-products. You should recognize the $ 2A-X $ to be a central reflection of X in $ A $, $ \sigma_A(X)$. The properly scaled displacement vector is nothing but the projection of $ X-A $ to the mirror line. Thus the geometrical meaning of formula (1) is here:


Geometrical interpretation of the formula for a reflection in a line.

2. Review of Some Vector Geometry

To see where the projection formula comes from, let’s recall some facts about vectors.

$F\frac{D}{|D|}=|F|\ cos\angle FD$

gives the magnitude of the projection of $F$ in the direction of $D$.

`F_D\ =\ (F frac{D}{|D|})frac{D}{|D|}\ =\ frac{F D}{D D}D`.

`F_D` is called the `D`-component of `F`. - The component of `F` in the direction perpendicular to `D` is given by the formula `F_D^{_|_}=F-F_D`.

We may denote this component by `F_D^{_|_}` and also by `F_{D^{_|_}}`. The first is the vector $ F_D $ turned $ 90^\circ $ to the right. The second is the component of $ F $ in the directions `D^{_|_}`. That these two vectors are the same requires a proof.

Proof. We must show `F_D^{_|_}=F_{D^{_|_}}`. Rather than prove this equality directly, it’s slightly easier to show that `F_D^{_|_}` has the same direction and magnitude as `F_{D^{_|_}}`.

Its direction must be perpendicular to `D`:

`(F-F_D) D\ =\ F D-F_D D\ =\ F D-(frac{F D}{D D}D D)\ =\ F D-F D\ =\ 0`.

Thus `F_D^{_|_}` is perpendicular to `D` and the component `F_D` so we may think of of `F_D^{_|_}` and `F_D` as the sides of a right triangle.

`(F-F_D)^2+F_D^2 = F^2`

3. Geometrical Definition of a Reflection

Let $ m $ be any line, called the mirror of the reflection $ \sigma_m $. Then for every point $ X $ on $ m $, set $ sigma_\m(X)=X $. Thus, every point on the line $ m $ is fixed by $ sigma_m $. Otherwise, for points not on $ m $, let $ X_m $ be the nearest point to $ X $ on $ m $.

Since we want $ X_\m $ to be the midpoint of the segment from $ X $ to $ X^\sigma $ we can solve the equation



Note that we have seen such a formula before. It says that $ X^\sigma = \sigma_{X_m} $.

Even though we have not given an explicit formula for $ X_m $, we can already prove that

Proof: By definition, to show $ \sigma = \sigma_m $ is an isometry we only need to show that it preserves distance: $ |X^\sigma-Y^\sigma|=|X-Y| $.

Let’s compute `X^sigma-Y^sigma=(2X_\m-X)-(2Y_\m-Y)=2(X_\m-Y_\m)-(X-Y)=2((X-Y)D)D-(X-Y)`.

To make the algebra more transparent, substitute $ Z = X-Y $ so that the right most expression of the above equation looks like this $ 2(ZD)D-Z $. Then to compute the length, we compute its square.

`|X^sigma-Y^sigma|^2=4(Z D)^2D^2-4(Z D)(Z D)+Z^2=4(Z D)^2-4(Z D)^2+Z^2=Z^2=|Z|^2`.

But `|X^sigma-Y^sigma|^2=|X-Y|^2` means `sigma` is an isometry.

$ \square $

4. Explicit Formula for a Reflection

We should still derive the explicit equation (1), although we will rarely, if ever, need to use it. We will rely on our geometric intuitions.

Question 1.

If we set $ F = X-A $ and $ D = \frac{B-A}{\|B-A\|}$ what part of the RHS of equation (1) is just $ F_D $ ?

Answer. [show] ▶

$ F_D = (FD)\ D = ((X-A) \frac{B-A}{\|B-A\|}) frac{B-A}{\|B-A\|} = \frac{(X-A)(B-A)}{(B-A)(B-A)}(B-A) $.

Question 2.

Why is $ X_\m = A + F_D $ ?

Answer. [show] ▶

Because $ \triangle AX_\m X $ is a right triangle.

Question 3.

How does this "finish" the derivations of (1)?

Answer. [show] ▶

$ X^\sigma = 2X_\m - X = 2A - X + 2F_D $ .

Question 4.

Can you draw a picture that makes $ X^\sigma = A + (A-X) + 2(X_\m - A) $ geometrically clear from the rules of similar triangles?

Answer. [show] ▶

The point $ B $ is not in the figure. It can be anywhere on $ \m $ since we only use it to find the orthogonal projection of $ F = X-A $ to $ F_\m = X_\m -A $. Note that when $ Y = A + (A-X) $ then $ A $ is the midpoint of $ \bar{YX} $. Finally, if the base vector is $ 2(X_\m - A) $, then $ X_\m $ is the midpoint of the vertical leg of this right triangle. Therefore the point $ X^\sigma $ is indeed the reflection of $ X $ in $ \m $.

5. Reflections and Fixed Points

Let us review what we know about fixed points and isometries.

Let’s see to what extent the fixed points of an isometry characterize the transformation. By "characterize" an isometry we mean that another isometry with the same characteristics must be the same isometry.

Proof of R1. Calculate $ |X-Q|=_1|X^\alpha - Q^\alpha|=_2|X^\alpha - Q| $ where the $ =_1 $ expresses the definition of an isometry and $ =_2 $ expresses the hypothesis of the assertion. In other words, $ Q $ is equidistant from the end points of the segment $ X X^\alpha$. But, as we have used many times before, the locus of points equidistant from the end points of a segment is the perpendicular bisector of the segment.

Proof of R2. Every point $ X $ on $ (PQ) $ has the form $ X=(1-t)P+tQ, t\in RR $. Relative to some third point $ K $ not on the line, this expresses $ X $ in terms of its barycentric coordinates relative to $ \triangle PQR $. By the Barytheorem and the hypothesis we have that $ X^\alpha = (1-t)P^\alpha + t Q^\alpha = (1-t)P + t Q = X$.

To actually identify $ \alpha $ as the reflection $ \sigma_\ell $, we again use some point $ K $ not on $ \ell $. Since $ \alpha \ne \iota $, we know that $ K $ cannot be a fixed point. Then by R1, both $ P, Q $ are on the $ perbis(K,K^\alpha)$, hence $ \ell $ is that perpendicular bisector. So $ K^\alpha = \sigma_\ell(K) $.

To show that $ \alpha = \sigma_\ell $ we consider $ \omega = \sigma_\ell \alpha $ and calculate the three non-collinear points, $ P,Q,K $ are all fixed by $ \omega $. Hence by the "tripod theorem" $ \omega = \iota $. Multiply both sides of this equation by $ \sigma_\ell $ yields the desired result. (Be sure you do this multiplication in the margin when you read this sentence!)

Proof of R3. Since $ Q $ is assumed to be the only fixed point, for any other point $ X^\alpha \ne X $ and by R1, $ Q $ in on $ m = perbis(X,X^\alpha) $. Now, note that $ \omega = \sigma_m \alpha $ has two fixed points, $ P and X $.

Now, $ \omega \ne \iota $ because if it were, then $ \alpha = \sigma_m $ which has infinitely many fixed points. Hence R2 applies, and $ \omega = sigma_\n $, where $ n=(QX) $. Multiplying both sides by the involution $ \sigma_m $ yields the desired $ \alpha = \sigma_m \sigma_\n $.

Proof of R4. We already know that translations are the product of two reflections in parallel mirrors, that are (both) perpendicular to the translation vector, and half its magnitude apart.

We also know that every isometry factors into the product of an osometry followed by a translation. By definition, an osometry fixes the origin. So, by what we have shown above, the osometry is either

6. First Conjugacy Theorem

We draw our first, fundamental conclusion from the properties of reflections, specifically from R2.

The conjugate $ \mu^\alpha $ has deep geometrical meaning, but an immediate algebraic meaning is the degree to which the two isometries do not commute.

Question 5.

Given two isometries $ \alpha,\mu $, for what isometry $ \xi $ is the equation $ \alpha \mu = \xi \alpha $ true?

Answer. [show] ▶

Multiply both sides of the equation "on the right" by $ \alpha^{-1}$, to get $ \alpha \mu \alpha^{-1} = \xi \alpha \alpha^{-1} = \xi $

The conjugate operation has a deeper algebraic meaning than just this. We note, in particular, that if $ \mu \ne \iota $ , neither is its conjugate. For suppose not, and $ \iota = \alpha \mu \alpha^{-1} $. Multiply both sides by $ \alpha^{-1} $ on the left, and by $ \alpha $ on the right and get a contradiction (do it!).

To conclude this section, we prove one profound theorem.

Proof: Abbreviate $ m := \ell^\alpha $. Since $ \ell^\alpha = \alpha(\ell) $, for all $ (Mm) $ there exists $ (L\ell) $ such that $ M = \alpha(L) $. We now apply the LHS of the equation to the point $ M $ .

$ \alpha \sigma_{\ell} \alpha^{-1}(M) = \alpha \sigma_{\ell}(L) = \alpha (L) = M $.

Thus $ \alpha \sigma_{\ell} \alpha^{-1} $ has every point on $ m $ as a fixed point, and not being the identity, it must be the reflection in that line.


This meaning of charm is reserved for this course only. We have thus found that the identity is not charming at all. Reflections have charm 1. Translations have charm 2. In the next section we will discover that rotations are equally charming as translations. We already know that no isometry is more charming than charm 4. Actually, we will discover that 3 is as charming as any isometries get. And the isometries with charm 3 are beautiful indeed.