Wessel’s Algebra and Morley’s Theorem

Abstract

This paper is devoted to provide a proof of F. Morley’s theorem concerning triangles in the Euclidean plane $\mathcal{E}$ (see Theorem 1 in the Introduction section) phrased in terms of the geometric algebra $\mathcal{G}$ of $\mathcal{E}$ (called Wessel’s algebra). This algebra is studied in detail in Section 2, its uses in describing isometries of $\mathcal{E}$ in Section 3, its bearing on the geometry of Morley’s construction in Section 4, and the claimed proof in Section 5. Morley’s theorem can be extended by using all the trisectors (interior and exterior) of a triangle, and suitable intersections of them. These intersections form what we call Morley’s constellation and out of it 36 generalized Morley triangles can be formed. Among these triangles, 27 are equilateral and with sides parallel to the original Morley triangle (Appendix B). The 36 triangles are depicted in Appendix C. All graphics in this work have been created by the author.

1. Introduction

Geometric algebra is a mathematical structure naturally associated with any quadratic linear space, and its main virtue is that it can be used to encode and intrinsically (i.e., in a coordinate-free way) develop the geometry of such spaces. Beyond its use in a rich variety of purely mathematical developments, a surprising range of applications have been brought to light in the last hundred years, particularly in the wake of the early breakthroughs of W. Pauli (1900–1958) and P.A.M. Dirac (1902–1984) to model the electron’s spin. Relevant developments in the literature have been gathered in ⋄15 in Appendix A.

Despite the formidable panorama outlined above, a geometric algebra proof of the following ever surprising theorem seems to be missing:

Theorem 1 

(Frank Morley, 1899). For any triangle $ABC$ in the Euclidean plane, its Morley triangle $XYZ$ (see Figure 1) is equilateral.

The situation is all the more surprising when a search of the literature reveals a sustained fascination of researchers with this theorem and also with some of its natural generalizations. We refer to ⋄16 in Appendix A for a brief guide to the references on these topics. Since the methods used in those works are specific to Euclid-style plane geometry, it is reasonable to expect potential advantages in exploring arguments expressed in a formalism, such as geometric algebra, that by its nature transcends those limitations, at least in the sense that whatever is learned from the geometric algebra of the Euclidean plane (which we call Wessel’s algebra) will greatly facilitate the study of geometric topics in any dimension, including spaces that are not necessarily Euclidean.

Yet, the question of who may be most interested in finding a proof of Theorem 1 using the geometric algebra formalism presents an auspicious opportunity to address readers eager to find a forthright introduction to Wessel algebra as a gateway to geometric algebra in general, and, as an extra reward, its use in producing a novel proof of Theorem 1 as well as connections to its natural generalizations.

With these premises in mind, the paper is structured as follows: Wessel’s algebra is presented in Section 2, with emphasis on how isometries (translations, rotations and reflections) are handled; Section 3 is devoted to providing explicit expressions for the various compositions of symmetries and Section 4 to establishing a number of symmetry properties of the Morley configuration depicted in Figure 1, while the promised proof of Theorem 1 is laid out in Section 5.

The paper has three appendices: Appendix A, to accommodate annotations to a few statements in the preceding sections; Appendix B, in which the definitions of the angle trisectors of a triangle are defined, the constellation of points obtained as intersections of pairs of trisectors of the same ‘color’ is depicted, and its surprising properties are described; and Appendix C, devoted to drawing the 27 equilateral triangles that can be formed with vertices in Morley’s constellation as well as the nine Morley triangles that are not equilateral.

2. The Wessel Algebra

In the approach to plane Euclidean geometry with geometric algebra the platform is the oriented Euclidean plane $\mathcal{E}$ (for a recap of this notion, see ⋄17 in Appendix A), and the finesse is provided by its associated geometric algebra. Despite appearances, this arrangement differs fundamentally from the traditional modeling of both the algebra and the Euclidean plane in the formalism of the usual (algebraic) complex numbers $\mathbf{C}$ (the resulting system is usually known as Argand’s plane). To appraise the sharp contrast between the two approaches in more detail, see ⋄18 in Appendix A. Here, it is just remarked that if we identify $\mathcal{E}$ with $\mathbf{C}$, then i commutes with vectors, which is a major difference from the fact that the geometric entity corresponding to i (which is to be denoted by $\mathfrak{i}$) anticommutes with vectors. Of course, this calls for some care, but this fact, rather than a hurdle, is actually one of the assets which confer $\mathbb{C}=\{a+b\mathfrak{i}|a,b\in \mathbb{R}\}$ its resilience in the face of geometric problems in the Euclidean plane.

• ⋄1 (Plane Euclidean geometry and geometric algebra). An expedient way to describe this algebra is to start with Grassmann’s exterior algebra of $\mathcal{E}$, $(\mathcal{G},\wedge )$, and enrich it with its geometric product. Let us spell this out in detail.

We have $\mathcal{G}={\mathcal{G}}^0\oplus {\mathcal{G}}^1\oplus {\mathcal{G}}^2$, where ${\mathcal{G}}^k$ is the space of k-vectors: ${\mathcal{G}}^0=\mathbb{R}$ (0-vectors or scalars, denoted by Greek symbols), ${\mathcal{G}}^1=\mathcal{E}$ (1-vectors or simply vectors, denoted by Latin symbols like $u,v,\dots$) and ${\mathcal{G}}^2$ (2-vectors or bivectors, typically $v\wedge v^{\prime }$, $v,v^{\prime }\in \mathcal{E}$). The bivector $v\wedge v^{\prime }$ is an algebraic representation of the area element corresponding to the oriented parallelogram $[v,v^{\prime }]=\{\lambda v+{\lambda }^{\prime }{v}^{\prime }|\lambda ,{\lambda }^{\prime }\in [0,1]\}$. The wedge product ∧ is bilinear and $v\wedge v=0$, indicating that the area element of $[v,v]$ is 0. From $(v+v^{\prime })\wedge (v+v^{\prime })=0$, we conclude that $v^{\prime }\wedge v=-v\wedge v^{\prime }$, so that the area element of $[v^{\prime },v]$ is opposite of the area element of $[v,v^{\prime }]$.

If we choose any basis of $\mathcal{E}$, say $\mathit{e}=e_1,e_2$, and use the decompositions $v=\alpha e_1+\beta e_2$ and $v^{\prime }={\alpha }^{\prime }{e}_1+{\beta }^{\prime }{e}_2$, we immediately get $v\wedge v^{\prime }=(\alpha {\beta }^{\prime }-{\alpha }^{\prime }\beta )e_1\wedge e_2={\det }_{\mathit{e}}(v,v^{\prime })e_1\wedge e_2$. This shows that $\{1,e_1,e_2,e_1\wedge e_2\}$ is a basis of $\mathcal{G}$. Note that $v\wedge v^{\prime }=e_1\wedge e_2$ if and only if ${\det }_{\mathit{e}}(v,v^{\prime })=1$, a condition that happens when $\{v,v^{\prime }\}$ is a positively oriented orthonormal basis of $\mathcal{E}$ (as in this case $v=e_1\cos \phi +e_2\sin \phi ,v^{\prime }=-e_1\sin \phi +e_2\cos \phi$ for some $\phi$). In particular the bivector $\mathfrak{i}=e_1\wedge e_2$ is the same for all positively oriented orthonormal bases and it is usually called the pseudoscalar of $\mathcal{E}$ (it is the unit area element). Thus we have ${\mathcal{G}}^2=\mathbb{R}\mathfrak{i}$ and elements $x\in \mathcal{G}$ (which are called multivectors) have the form

$$x=\alpha +v+\beta \mathfrak{i}(\alpha ,\beta \in \mathbb{R},v\in \mathcal{E}).$$

• ⋄2 (The geometric product). There is a unique bilinear associative unital product $x{x}^{\prime }\in \mathcal{G}$ ($x,x^{\prime }\in \mathcal{G}$), such that

  $$v{v}^{\prime }=v·v^{\prime }+v\wedge v^{\prime }$$

  (1)

  for all $v,v^{\prime }\in \mathcal{E}$. This product is called geometric product. Observe that (1) implies that $v^2=v·v$, and $v^{\prime }v=-v{v}^{\prime }$ if (and only if) $v·v^{\prime }=0$ (denoted by $v\perp v^{\prime }$, or, in words, if and only if v and $v^{\prime }$ are orthogonal). We see, in particular, that any non-zero vector v is invertible: $v^{-1}=\frac{v}{v·v}=\frac{v}{{\left|v\right|}^2}$. This feature is paramount in geometric algebra, as computations in it have the flavor of computations in $\mathbb{R}$, except that the geometric product is not commutative.

Uniqueness of the geometric product follows from checking that the conditions in the statement determine the multiplication table of the basis $\{1,e_1,e_2,\mathfrak{i}\}$ of $\mathcal{G}$ corresponding to an orthonormal basis $\mathit{e}=e_1,e_2$ of $\mathcal{E}$, as shown in

Table 1. Geometric products of the basis elements of $\mathcal{G}$. In the form $1,e_1,e_2,\mathfrak{i}$ (Left) and in the form $e_{00},e_{10},e_{01},e_{11}$ (Middle). Right: The transposition counter t: its value is 1 if $j_2=k_1=1$ (indicating that in the product $e_2$ appears before $e_1$) and 0 otherwise.

1

$e_1$

$e_2$

$\mathfrak{i}$

$e_{00}$

$e_{10}$

$e_{01}$

$e_{11}$

t

10

01

11

$e_1$

1

$\mathfrak{i}$

$e_2$

$e_{10}$

$e_{00}$

$e_{11}$

$e_{01}$

10

$+1$

$+1$

$+1$

$e_2$

$-\mathfrak{i}$

1

$-e_1$

$e_{01}$

$-e_{11}$

$e_{00}$

$-e_{10}$

01

$-1$

$+1$

$-1$

$\mathfrak{i}$

$-e_2$

$e_1$

$-1$

$e_{11}$

$-e_{01}$

$e_{10}$

$-e_{00}$

11

$-1$

$+1$

$-1$

(Left). For example $e_1{e}_2=e_1\wedge e_2=\mathfrak{i}$, because $e_1·e_2=0$; $e_1\mathfrak{i}=e_1{e}_1{e}_2=e_2$, as $e_1^2=e_1·e_1=1$; ${\mathfrak{i}}^2=e_1{e}_2{e}_1{e}_2=-e_1^2{e}_2^2=-1$, as $e_2{e}_1=-e_1{e}_2$; and so on.

Existence of the geometric product follows by defining a bilinear product by means of Table 1 (Left) and checking that it satisfies (1) and is associative. The relation (1) is straightforward: if $v=\alpha e_1+\beta e_2$ and $v^{\prime }={\alpha }^{\prime }{e}_1+{\beta }^{\prime }{e}_2$, then using bilinearity and the values provided by Table 1 (Left) we have:

$$v{v}^{\prime }=(\alpha e_1+\beta e_2)({\alpha }^{\prime }{e}_1+{\beta }^{\prime }{e}_2)=\alpha {\alpha }^{\prime }+\beta {\beta }^{\prime }+(\alpha {\beta }^{\prime }-\beta {\alpha }^{\prime })\mathfrak{i}=v·v^{\prime }+v\wedge v^{\prime }.$$

Notice the relation $v\wedge v=A(v,v^{\prime })\mathfrak{i}$ between the area element $v\wedge v^{\prime }$ and the oriented area $A(v,v^{\prime })=\alpha {\beta }^{\prime }-\beta {\alpha }^{\prime }$ of the parallelogram $[v,v^{\prime }]$.

It remains to prove that the geometric product is associative. This could be achieved by inspecting all possible cases that can be formed with Table 1, Left, as for example $\left(e_2\mathfrak{i}\right)e_1=e_2\left(\mathfrak{i}{e}_1\right)$, where the value of both sides is $-1$. But it is more enlightening to provide a conceptual argument, which here will be no more than accommodating to this case, the general proof presented in ([1], Sect. 1) by profiting from the particular features of $\mathcal{E}$. By writing $e_{00}=1,e_{10}=e_1,e_{01}=e_2,e_{11}=\mathfrak{i}$, the product table becomes Table 1, Middle, which can be summarized, regarding the subindices as ordered pairs of bits, by the formula

$$e_j{e}_k={(-1)}^{t(j,k)}{e}_{j+k},$$

(2)

where $j+k$ denotes the bit-wise sum of j and k (example: $01+11=10$), while $t(j,k)=1$ if $j_2=k_1=1$ and 0 otherwise (see Table 1, Right).

To check associativity, it suffices to see that

$$\left(e_j{e}_k\right)e_l=e_j\left(e_k{e}_l\right)\mathrm{for}\mathrm{all}j,k,l\in B=\{10,01,11\}.$$

Using (2), we have:

$$\begin{array}{cc}\hfill \left(e_j{e}_k\right)e_l& ={(-1)}^{t(j,k)}{e}_{j+k}{e}_l={(-1)}^{t(j,k)+t(j+k,l)}{e}_{(j+k)+l}\hfill \\ \hfill e_j\left(e_k{e}_l\right)& ={(-1)}^{t(k,l)}{e}_j{e}_{k+l}={(-1)}^{t(k,l)+t(j,k+l)}{e}_{j+(k+l)}.\hfill \end{array}$$

Since $(j+k)+l=j+(k+l)$, to conclude it is enough to see that

$$t(j,k)+t(j+k,l)\mathrm{and}t(k,l)+t(j,k+l)$$

have the same parity for all $j,k,l\in B$. Actually we claim that both sums have the same parity as $t(j,k)+t(j,l)+t(k,l)$, and this is clearly enough to complete the proof.

The claim itself follows from the fact that $t(j+k,l)$ and $t(j,k+l)$ have the same parity as $t(j,l)+t(k,l)$ and $t(j,k)+t(j,l)$, respectively. Since the analysis of both expressions is similar, let us focus on the details of $t(j+k,l)$. If j and k are disjoint (that is, they do not share a 1 in any position, which is equivalent to saying that the bit-wise product $jk$ is 00), then we actually have the equality $t(j+k,l)=t(j,l)+t(k,l)$, as follows from the definition of the transposition counter. In general, observing that $j+jk$, $k+jk$ and $jk$ are pair-wise disjoint (for example, $(j+jk)\left(jk\right)=jk+jk=00$, as the bit-wise product is distributive and associative and $jj=j$, $j+j=00$ for any j), we have

$$\begin{array}{cc}\hfill & t(j,l)=t(j+jk,l)+t(jk,l),\hfill \\ \hfill & t(k,l)=t(k+jk,l)+t(jk,l),\hfill \\ \hfill & t(j,l)+t(k,l)=t(j+jk,l)+t(k+jk,l)+2t(jk,l)\hfill \\ \hfill & =t(j+jk+k+jk)+2t(jk,l)\hfill \\ \hfill & =t(j+k)+2t(jk,l).\hfill \end{array}$$

• ⋄3 (Geometric algebra). The geometric algebra of $\mathcal{E}$ is $\mathcal{G}$ enriched with the geometric product, and we will use the same symbol $\mathcal{G}$ to denote it. For historical reasons (see [2]), we say that $\mathcal{G}$ is the $Wessel algebra$ of $\mathcal{E}$. The even subalgebra of $\mathcal{G}$ is

  $${\mathcal{G}}^{+}={\mathcal{G}}^0\oplus {\mathcal{G}}^2=\mathbb{R}\oplus \mathbb{R}\mathfrak{i}=\mathbb{C}.$$

  Since ${\mathfrak{i}}^2=-1$, $\mathbb{C}$ is isomorphic to the complex numbers $\mathbf{C}$, $\alpha +\beta \mathfrak{i}\mapsto \alpha +\beta i$, but the geometric nature of $\mathfrak{i}$ (the unit area element) justifies saying that $\mathbb{C}$ is the algebra (actually a field) of geometric complex numbers, and so we may conveniently refer to $\mathbf{C}$ as the field of algebraic complex numbers. Let us just remark that fields of geometric complex numbers are abundant in higher dimensions because there are plenty of bivectors $\mathfrak{j}$ such that ${\mathfrak{j}}^2=-1$ (for example $\mathfrak{j}=u{u}^{\prime }$, where u and $u^{\prime }$ are orthogonal vectors such that $u^2={u^{\prime }}^2=1$).

Table 1, Left, shows that $\mathfrak{i}$ anticommutes with vectors, and that $v\mathfrak{i}\in \mathcal{E}$ for all $v\in \mathcal{E}$. Actually, the map $\mathcal{E}\to \mathcal{E}$, $v\mapsto v\mathfrak{i}$, is a rotation of amplitude $\pi /2$. Indeed, for any unit vector u, let $u^{\perp }$ be the result of rotating u through $\pi /2$, so that $\{u,u^{\perp }\}$ is a positive orthonormal basis, and hence $\mathfrak{i}=u{u}^{\perp }$, which yields, on multiplying by u on the left, $u\mathfrak{i}=u^{\perp }$. In general, $v{e}^{\mathfrak{i}\phi }$ is the rotation of vector v through $\phi$ (see Figure 2):

$$v{e}^{\mathfrak{i}\phi }=v\cos \phi +v\mathfrak{i}\sin \phi =v\cos \phi +v^{\perp }\sin \phi .$$

(3)

Since $v{e}^{\mathfrak{i}\phi }=e^{-\mathfrak{i}\phi }v$ for any vector v, we also have the relation

$$v{e}^{\mathfrak{i}\phi }=e^{-\mathfrak{i}\phi /2}v{e}^{\mathfrak{i}\phi /2}.$$

(4)

This identity will be handy for finding expressions of the composition of symmetries and rotations.

• ⋄4 (Notational conventions). We need to distinguish two pragmatic uses of vectors: one is to indicate a position, in which case they will be called points and will be denoted by capital Latin letters like A, B, C, P, Q, X, …; the other use is as translators of points, in which case they will be denoted by lowercase letters such as $u,u^{\prime },v,v^{\prime },\dots$ and used in expressions like $Q=P+v=v+P$, or, equivalently, $v=Q-P$. Even though expressions such as $P+Q$ make sense in the abstract, for after all they can be seen as a sum of two vectors, we will avoid such usage as a way to enforce and highlight those two roles. In graphic illustrations, points are represented by bullet dots and vectors by arrows. The helpfulness of these distinctions will be apparent in the description of rotations and symmetries below.

An isometry of $\mathcal{E}$ is a map $f:\mathcal{E}\to \mathcal{E}$ such that $\left|f\right(X)-f(Y\left)\right|=|X-Y|$ for all $X,Y\in \mathcal{E}$. Since the length of a nonzero vector is positive, an isometry is a 1-to-1 map. For example, the map $T_v:X\mapsto X+v$, which is called the translation of $\mathcal{E}$ by v, is an isometry, as $T_v\left(X\right)-T_v\left(Y\right)=(X+v)-(Y+v)=X-Y$.

• ⋄5 (Rotations and symmetries). Now we are ready to articulate rotations and symmetries of the Euclidean plane $\mathcal{E}$ by means of the Wessel algebra, and to recognize its benefits.
  Rotations of $\mathcal{E}$. The rotation of points about a point P through an angle $\phi$ will be denoted by $R_{P,\phi }$ and it is defined by the formula $R_{P,\phi }\left(X\right)=P+(X-P)e^{\mathfrak{i}\phi }$. Plainly, we are rotating the vector $X-P$ through $\phi$ (see Figure 2) and adding the result to P, so that, in particular, P is a fixed point. If $\phi =0$, $R_{P,\phi }=\mathrm{Id}$ (for which all points are fixed); otherwise P is the only fixed point, for $X=P+(X-P)e^{\mathfrak{i}\phi }$ says that $(X-P)=(X-P)e^{\mathfrak{i}\phi }=0$ and this implies $X-P=0$ if $\phi \ne 0$. In fact $R_{P,\phi }$ is an isometry, for $R_{P,\phi }\left(X^{\prime }\right)-R_{P,\phi }\left(X\right)=(X^{\prime }-X)e^{\mathfrak{i}\phi }$, which has the same length as $X^{\prime }-X$.
  Symmetries of $\mathcal{E}$. Take an arbitrary line ℓ, say the line though a point P with direction vector u: $\mathsf{\ell }={\mathsf{\ell }}_{P,u}=P+\langle u\rangle =\{P+\lambda u|\lambda \in \mathbb{R}\}.$ With the notations of Figure 3, $S_{\mathsf{\ell }}\left(X\right)=X-2v^{″}$, which is the geometric definition of $S_{\mathsf{\ell }}$.

Now we claim that in the Wessel algebra $S_{\mathsf{\ell }}$ has the following expression:

$$S_{\mathsf{\ell }}\left(X\right)=P+u(X-P)u^{-1}.$$

(5)

Indeed, write $X-P=v^{\prime }+v^{″}$, where $v^{\prime }\in \langle u\rangle$ and $v^{″}\in {\langle u\rangle }^{\perp }$, as depicted in Figure 3. Since $v^{\prime }$ and $v^{″}$ commute and anticommute with u, respectively, $u(X-P)u^{-1}=v^{\prime }-v^{″}$. Therefore,

$$P+u(X-P)u^{-1}=P+v^{\prime }-v^{″}=X-2v^{″}=S_{\mathsf{\ell }}\left(X\right).$$

(6)

In the description of $S_{\mathsf{\ell }}$ we have used an arbitrary point $P\in \mathsf{\ell }$ and an arbitrary direction vector u of ℓ, but it is worth remarking that the expression of $S_{\mathsf{\ell }}$ only depends on ℓ, not on those arbitrary choices, for moving P along ℓ or rescaling u does not change $v^{″}$. Note that the expression (5) readily shows that $S_{\mathsf{\ell }}$ is an isometry, for $S_{\mathsf{\ell }}\left(X^{\prime }\right)-S_{\mathsf{\ell }}\left(X\right)=u(X^{\prime }-X)u^{-1}$, and hence, its square length is $u(X^{\prime }-X)u^{-1}u(X^{\prime }-X)u^{-1}=u{(X^{\prime }-X)}^2{u}^{-1}={(X^{\prime }-X)}^2$ (u commutes with the scalar ${(X^{\prime }-X)}^2$).

It is also worth noting that a point X is fixed by $S_{\mathsf{\ell }}$ if and only if $X\in \mathsf{\ell }$, for such points are precisely those that satisfy $v^{″}=0$ (in the above notations) and this condition is equivalent to saying that $X\in \mathsf{\ell }$.

Given a vector $w\Vert \mathsf{\ell }$, the composition $T_w{S}_{\mathsf{\ell }}$ will be denoted by $S_{\mathsf{\ell },w}$ (or $S_{P,u,w}$ if ℓ is specified as $\mathsf{\ell }=P+\langle u\rangle$). It is an isometry of $\mathcal{E}$ and it has no fixed points if $w\ne 0$, for the relation $X=w+X-2v^{″}$ implies $w=2v^{″}$ and this can only happen, as w and $v^{″}$ are perpendicular, when $w=v^{″}=0$. The isometries of the form $S_{\mathsf{\ell },w}$ ($w \Vert \mathsf{\ell }$) are called gliding symmetries.

3. Composition of Translations, Rotations and Symmetries

Having expressions in the Wessel algebra for rotations and symmetries, we are ready to explore the compositions specified in

Table 2. In each cell we have the composition of the isometry on top of its column followed by the isometry at the head of its row, but we omit the composition symbol $\circ$. The case $T_{v^{\prime }}{T}_v=T_{v+v^{\prime }}$ is obvious. All other cases require closer attention.

$\circ$	$T_v$	$R_{P,\phi }$	$S_{\mathsf{\ell }}$
$T_{v^{\prime }}$	$T_{v^{\prime }}{T}_v$	$T_{v^{\prime }}{R}_{P,\phi }$	$T_{v^{\prime }}{S}_{\mathsf{\ell }}$
$R_{P^{\prime },{\phi }^{\prime }}$	$R_{P^{\prime },{\phi }^{\prime }}{T}_v$	$R_{P^{\prime },{\phi }^{\prime }}{R}_{P,\phi }$	$R_{P^{\prime },{\phi }^{\prime }}{S}_{\mathsf{\ell }}$
$S_{{\mathsf{\ell }}^{\prime }}$	$S_{{\mathsf{\ell }}^{\prime }}{T}_v$	$S_{{\mathsf{\ell }}^{\prime }}{R}_{P,\phi }$	$S_{{\mathsf{\ell }}^{\prime }}{S}_{\mathsf{\ell }}$

. To ease notations, we are going to write $u_{\phi }=e^{\mathfrak{i}\phi }$, and use the ensuing relations $u_{\phi }^{-1}=e^{-\mathfrak{i}\phi }=u_{-\phi }$ and $u_{\phi }{u}_{{\phi }^{\prime }}=u_{\phi +{\phi }^{\prime }}$.

• ⋄6 ($f=T_{v^{\prime }}{R}_{P,\phi }$, $\phi \ne 0$, and $f^{\prime }=R_{P^{\prime },{\phi }^{\prime }}{T}_v$, ${\phi }^{\prime }\ne 0$). If we let $P^{\prime }=P+v^{\prime }{(1-u_{\phi })}^{-1}$, then

  $$T_{v^{\prime }}{R}_{P,\phi }=R_{P^{\prime },\phi }.$$

  (7)

  First let us solve for X the equation $X=f\left(X\right)=v^{\prime }+P+(X-P)u_{\phi }$: $(X-P)(1-u_{\phi })=v^{\prime }$, or $X=P+v^{\prime }{(1-u_{\phi })}^{-1}$. This shows that $P^{\prime }$ in the statement is the only fixed point of f. Now it suffices to check that $R_{P^{\prime },\phi }\left(X\right)=f\left(X\right)$ for any point $X\in \mathcal{E}$:

  $$\begin{array}{cc}\hfill R_{P^{\prime },\phi }\left(X\right)& =P^{\prime }+(X-P^{\prime })u_{\phi }\hfill \\ \hfill & =P+v^{\prime }{(1-u_{\phi })}^{-1}+(X-P-v^{\prime }{(1-u_{\phi })}^{-1})u_{\phi }\hfill \\ \hfill & =P+(X-P)u_{\phi }+v^{\prime }{(1-u_{\phi })}^{-1}-v^{\prime }{(1-u_{\phi })}^{-1}{u}_{\phi }\hfill \\ \hfill & =P+(X-P)u_{\phi }+v^{\prime }(1-u_{\phi }){(1-u_{\phi })}^{-1}\hfill \\ \hfill & =P+(X-P)u_{\phi }+v^{\prime }=f\left(X\right).\hfill \end{array}$$

Remark 1. 

The expression ${(1-u_{\phi })}^{-1}$ appearing in the definition of $P^{\prime }$ is equal to $u_{\frac{\pi }{2}-\frac{\phi }{2}}/2\sin \frac{\phi }{2}$, so that it scales by $\frac{1}{2\sin \phi /2}$ the result of rotating $v^{\prime }$ through $\frac{\pi }{2}-\frac{\phi }{2}$. Indeed,

$$\begin{array}{cc}\hfill 1-u_{\phi }& =1-\cos \phi -\mathfrak{i}\sin \phi \hfill \\ \hfill & =2{\sin }^2\frac{\phi }{2}-2\mathfrak{i}\sin \frac{\phi }{2}\cos \frac{\phi }{2}\hfill \\ \hfill & =2\sin \frac{\phi }{2}(\sin \frac{\phi }{2}-\mathfrak{i}\cos \frac{\phi }{2})\hfill \\ \hfill & =2\sin \frac{\phi }{2}\left(\cos (\frac{\pi }{2}-\frac{\phi }{2})-\mathfrak{i}\sin (\frac{\pi }{2}-\frac{\phi }{2})\right)\hfill \\ \hfill & =2\sin \frac{\phi }{2}{u}_{\frac{\phi }{2}-\frac{\pi }{2}},\hfill \end{array}$$

which proves the claimed expression.

In the case $f^{\prime }=R_{P^{\prime },{\phi }^{\prime }}{T}_v$, ${\phi }^{\prime }\ne 0$, if we let $P=P^{\prime }+v{u}_{{\phi }^{\prime }}{(1-u_{{\phi }^{\prime }})}^{-1}$, then

$$R_{P^{\prime },{\phi }^{\prime }}{T}_v=R_{P,{\phi }^{\prime }}.$$

(8)

As in the previous case, P turns out to be the only fixed point of $f^{\prime }$ and the relation $R_{P,{\phi }^{\prime }}\left(X\right)=f^{\prime }\left(X\right)$ is obtained by a similar computation.

Remark 2. 

Using the previous Remark, we get

$$u_{{\phi }^{\prime }}{(1-u_{{\phi }^{\prime }})}^{-1}=u_{{\phi }^{\prime }}{u}_{\frac{\pi }{2}-\frac{{\phi }^{\prime }}{2}}/2\sin \left(\frac{{\phi }^{\prime }}{2}\right)=u_{\frac{\pi }{2}+\frac{{\phi }^{\prime }}{2}}/2\sin \left(\frac{{\phi }^{\prime }}{2}\right).$$

• ⋄7 ($f=R_{P^{\prime },{\phi }^{\prime }}{R}_{P,\phi }$). We have

  $$\begin{array}{cc}\hfill f\left(X\right)& =R_{P^{\prime },{\phi }^{\prime }}(P+(X-P)u_{\phi })\hfill \\ \hfill & =P^{\prime }+\left(P-P^{\prime }+(X-P)u_{\phi }\right)u_{{\phi }^{\prime }}\hfill \\ \hfill & =P^{\prime }-(P^{\prime }-P)u_{{\phi }^{\prime }}+(X-P)u_{\phi }{u}_{{\phi }^{\prime }}\hfill \\ \hfill & =P+(P^{\prime }-P)(1-u_{{\phi }^{\prime }})+(X-P)u_{\phi }{u}_{{\phi }^{\prime }}.\hfill \end{array}$$

  If $u_{\phi }{u}_{{\phi }^{\prime }}=1$, this equation says that $f\left(X\right)=X+(P^{\prime }-P)(1-u_{{\phi }^{\prime }})=X+v$, where $v=(P^{\prime }-P)(1-u_{{\phi }^{\prime }})$, and this means that $f=T_v$.

If $u_{\phi }{u}_{{\phi }^{\prime }}\ne 1$, then f has a unique fixed point $P^{″}=P+(P^{\prime }-P)\frac{1-u_{{\phi }^{\prime }}}{1-u_{\phi }{u}_{{\phi }^{\prime }}}$, for the equation $X=f\left(X\right)$ is equivalent to $(X-P)(1-u_{\phi }{u}_{{\phi }^{\prime }})=(P^{\prime }-P)(1-u_{{\phi }^{\prime }})$ and $X=P^{″}$ is its unique solution. Moreover, a short computation shows that in this case $f=R_{P^{″},\phi +{\phi }^{\prime }}$:

$$\begin{array}{cc}\hfill R_{P^{″},\phi +{\phi }^{\prime }}\left(X\right)& =P^{″}+(X-P^{″})u_{\phi }{u}_{{\phi }^{\prime }}\hfill \\ \hfill & =P+(P^{\prime }-P)\frac{1-u_{{\phi }^{\prime }}}{1-u_{\phi }{u}_{{\phi }^{\prime }}}+\left(X-P-(P^{\prime }-P)\frac{1-u_{{\phi }^{\prime }}}{1-u_{\phi }{u}_{{\phi }^{\prime }}}\right)u_{\phi }{u}_{{\phi }^{\prime }}\hfill \\ \hfill & =P+(P^{\prime }-P)(1-u_{{\phi }^{\prime }})+(X-P)u_{\phi }{u}_{{\phi }^{\prime }}\hfill \end{array}$$

which agrees with $f\left(X\right)$.

• ⋄8 ($f=T_{v^{\prime }}{S}_{\mathsf{\ell }}$ and $f^{\prime }=S_{{\mathsf{\ell }}^{\prime }}{T}_v$). The composition f is a gliding symmetry: if $v^{\prime }=v_{\mathsf{\ell }}^{\prime }+v_{\perp }^{\prime }$, where $v_{\mathsf{\ell }}^{\prime }\Vert \mathsf{\ell }$, $v_{\perp }^{\prime }\perp \mathsf{\ell }$, and ${\mathsf{\ell }}^{\prime }=\mathsf{\ell }+\frac{1}{2}{v}_{\perp }^{\prime }$, then

  $$T_{v^{\prime }}{S}_{\mathsf{\ell }}=S_{{\mathsf{\ell }}^{\prime },v_{\mathsf{\ell }}^{\prime }}.$$

  (9)

  This is a symmetry if and only if $v_{\mathsf{\ell }}^{\prime }=0$, that is, if and only if $v^{\prime }\perp \mathsf{\ell }$.

Indeed, in any case $T_{v^{\prime }}{S}_{\mathsf{\ell }}=T_{v_{\mathsf{\ell }}^{\prime }}{T}_{v_{\perp }^{\prime }}{S}_{\mathsf{\ell }}$, so it suffices to see that $T_{v_{\perp }^{\prime }}{S}_{\mathsf{\ell }}=S_{{\mathsf{\ell }}^{\prime }}$. Using a representation $\mathsf{\ell }={\mathsf{\ell }}_{P,u}$, we have ${\mathsf{\ell }}^{\prime }={\mathsf{\ell }}_{P+\frac{1}{2}{v}_{\perp }^{\prime },u}$. Hence,

$$\begin{array}{cc}\hfill S_{{\mathsf{\ell }}^{\prime }}\left(X\right)& =P+\frac{1}{2}{v}_{\perp }^{\prime }+u(X-P-\frac{1}{2}{v}_{\perp }^{\prime })u^{-1}\hfill \\ \hfill & =v_{\perp }^{\prime }+P+u(X-P)u^{-1}\hfill \\ \hfill & =v_{\perp }^{\prime }+S_{\mathsf{\ell }}\left(X\right)=T_{v_{\perp }^{\prime }}\left(S_{\mathsf{\ell }}\left(X\right)\right).\hfill \end{array}$$

where in the second step we use that u and $v_{\perp }^{\prime }$ anticommute.

The composition $f^{\prime }$ is also a gliding symmetry: if $v=v_{{\mathsf{\ell }}^{\prime }}+v_{\perp }$, where $v_{{\mathsf{\ell }}^{\prime }}\Vert {\mathsf{\ell }}^{\prime }$, $v_{\perp }\perp {\mathsf{\ell }}^{\prime }$, and $\mathsf{\ell }={\mathsf{\ell }}^{\prime }-\frac{1}{2}{v}_{\perp }$, then

$$S_{{\mathsf{\ell }}^{\prime }}{T}_v=T_{v_{{\mathsf{\ell }}^{\prime }}}{S}_{\mathsf{\ell }}=S_{\mathsf{\ell },v_{{\mathsf{\ell }}^{\prime }}}.$$

(10)

This is a symmetry if and only if $v_{{\mathsf{\ell }}^{\prime }}=0$, that is, if and only if $v\perp {\mathsf{\ell }}^{\prime }$.

Indeed, if ${\mathsf{\ell }}^{\prime }={\mathsf{\ell }}_{P^{\prime },u^{\prime }}$ and $P=P^{\prime }-\frac{1}{2}{v}_{\perp }$, then $\mathsf{\ell }={\mathsf{\ell }}_{P,u^{\prime }}$ and

$$\begin{array}{cc}\hfill S_{{\mathsf{\ell }}^{\prime }}\left(T_v\left(X\right)\right)& =S_{{\mathsf{\ell }}^{\prime }}(X+v_{{\mathsf{\ell }}^{\prime }}+v_{\perp })\hfill \\ \hfill & =P^{\prime }+u^{\prime }(X-P^{\prime }+v_{{\mathsf{\ell }}^{\prime }}+v_{\perp }){u^{\prime }}^{-1}\hfill \\ \hfill & =v_{{\mathsf{\ell }}^{\prime }}+P^{\prime }+u^{\prime }(X-P+v_{\perp }){u^{\prime }}^{-1}\hfill \\ \hfill & =v_{{\mathsf{\ell }}^{\prime }}+P^{\prime }-\frac{1}{2}{v}_{\perp }+u^{\prime }(X-P^{\prime }+\frac{1}{2}{v}_{\perp }){u^{\prime }}^{-1}\hfill \\ \hfill & =v_{{\mathsf{\ell }}^{\prime }}+P+u^{\prime }(X-P){u^{\prime }}^{-1}\hfill \\ \hfill & =v_{{\mathsf{\ell }}^{\prime }}+S_{\mathsf{\ell }}\left(X\right)=S_{\mathsf{\ell },v_{{\mathsf{\ell }}^{\prime }}}\left(X\right).\hfill \end{array}$$

In the third step, use that $u^{\prime }{v}_{{\mathsf{\ell }}^{\prime }}{u^{\prime }}^{-1}=v_{{\mathsf{\ell }}^{\prime }}$ (for $v_{{\mathsf{\ell }}^{\prime }}$ commutes with $u^{\prime }$) and in the fourth that $u^{\prime }{v}_{\perp }{u^{\prime }}^{-1}=-v_{\perp }$ (for $v_{\perp }$ anticommutes with $u^{\prime }$).

• ⋄9 ($f=R_{P^{\prime },{\phi }^{\prime }}{S}_{\mathsf{\ell }}$ and $f^{\prime }=S_{{\mathsf{\ell }}^{\prime }}{R}_{P,\phi }$). These compositions produce gliding symmetries. To deal with f (see Figure 4, Left), choose any representation ${\mathsf{\ell }}_{P,u}$ of ℓ and set $v=P^{\prime }-P$, $v^{\prime }=v{u}_{{\phi }^{\prime }}$ and $w=v-v^{\prime }$. Then

  $$\begin{array}{cc}\hfill R_{P^{\prime },{\phi }^{\prime }}{S}_{\mathsf{\ell }}\left(X\right)& =R_{P^{\prime },{\phi }^{\prime }}(P+u(X-P)u^{-1})\hfill \\ \hfill & =P^{\prime }+(P-P^{\prime }+u(X-P)u^{-1})u_{{\phi }^{\prime }}\hfill \\ \hfill & =P+(P^{\prime }-P)(1-u_{{\phi }^{\prime }})+u(X-P)u^{-1}{u}_{{\phi }^{\prime }}\hfill \\ \hfill & =P+(P^{\prime }-P)(1-u_{{\phi }^{\prime }})+u{u}_{{\phi }^{\prime }/2}(X-P)u_{{\phi }^{\prime }/2}^{-1}{u}^{-1}\hfill \\ \hfill & =w+S_{\overline{\mathsf{\ell }}}\left(X\right)\left[w=(P^{\prime }-P)(1-u_{{\phi }^{\prime }}),\overline{\mathsf{\ell }}={\mathsf{\ell }}_{P,\overline{u}},\overline{u}=u{u}_{{\phi }^{\prime }/2}\right]\hfill \\ \hfill & =w^{\prime }+S_{\overline{\mathsf{\ell }}+w^{″}/2}\left(X\right),w=w^{\prime }+w^{″},w^{\prime }\Vert \overline{\mathsf{\ell }},w^{″}\perp \overline{\mathsf{\ell }}.\hfill \end{array}$$

  In the third step use $P^{\prime }=P+(P^{\prime }-P)$ and in the fourth use Equation (4) to get

  $$u(X-P)u^{-1}{u}_{{\phi }^{\prime }}=u(X-P)u_{-{\phi }^{\prime }}{u}^{-1}=u{u}_{{\phi }^{\prime }/2}(X-P)u_{{\phi }^{\prime }/2}^{-1}{u}^{-1}.$$

  The last step relies on ⋄8.

Now consider $f^{\prime }=S_{{\mathsf{\ell }}^{\prime }}{R}_{P,\phi }$ (see Figure 4, Right). Choose any representation ${\mathsf{\ell }}^{\prime }={\mathsf{\ell }}_{P^{\prime },u^{\prime }}$. Let $Q=S_{{\mathsf{\ell }}^{\prime }}\left(P\right)$, $\overline{u}=u^{\prime }{u}_{-\phi /2}$, $v=Q-P=v^{\prime }+v^{″}$, with $v^{\prime }\Vert \overline{u}$ and $v^{″}\perp \overline{u}$, and $\overline{P}=Q-\frac{1}{2}{v}^{″}$. Then $f^{\prime }=S_{\overline{P},\overline{u},v^{\prime }}=S_{\overline{\mathsf{\ell }},v^{\prime }}$. Indeed:

$$\begin{array}{cc}\hfill f^{\prime }\left(X\right)& =S_{{\mathsf{\ell }}^{\prime }}(P+(X-P)u_{\phi })\hfill \\ \hfill & =P^{\prime }+u^{\prime }(P-P^{\prime }+(X-P)u_{\phi }){u^{\prime }}^{-1}\hfill \\ \hfill & =P^{\prime }+u^{\prime }(P-P^{\prime }){u^{\prime }}^{-1}+u^{\prime }(X-P)u_{\phi }{u^{\prime }}^{-1}\hfill \\ \hfill & =S_{{\mathsf{\ell }}^{\prime }}\left(P\right)+u^{\prime }{u}_{-\phi /2}(X-P)u_{\phi /2}{u^{\prime }}^{-1}\hfill \\ \hfill & =Q+\overline{u}(X-P){\overline{u}}^{-1}\hfill \\ \hfill & =\overline{P}+\frac{1}{2}{v}^{″}+\overline{u}(X-\overline{P}-\frac{1}{2}{v}^{″}){\overline{u}}^{-1}+\overline{u}(Q-P){\overline{u}}^{-1}\hfill \\ \hfill & =\overline{P}+\overline{u}(X-\overline{P}){\overline{u}}^{-1}+v^{\prime }\left[\overline{u}(Q-P){\overline{u}}^{-1}=\overline{u}v{\overline{u}}^{-1}=v^{\prime }-v^{″}\right]\hfill \\ \hfill & =v^{\prime }+S_{\overline{P},\overline{u}}\left(X\right)=S_{\overline{P},\overline{u},v^{\prime }}\left(X\right).\hfill \end{array}$$

• ⋄10 ($S_{{\mathsf{\ell }}^{\prime }}{S}_{\mathsf{\ell }}$). If $\mathsf{\ell }\nparallel {\mathsf{\ell }}^{\prime }$, this composition is the rotation $R_{P,2\phi }$ (Figure 5, Middle), where $P=\mathsf{\ell }\cap {\mathsf{\ell }}^{\prime }$ and $\phi =\phi (\mathsf{\ell },{\mathsf{\ell }}^{\prime })$ is the oriented angle from ℓ to ${\mathsf{\ell }}^{\prime }$ (Figure 5, Left). Otherwise, it is the translation $T_{2v^{″}}$, where $v^{″}$ is the vector orthogonal to ℓ (hence also to ${\mathsf{\ell }}^{\prime }$) such that $T_{v^{″}}(\mathsf{\ell })={\mathsf{\ell }}^{\prime }$ (Figure 5, Right).

Indeed, choosing unit direction vectors u and $u^{\prime }$ for ℓ and ${\mathsf{\ell }}^{\prime }$ such that $u·u^{\prime }=\cos \phi$, we have:

$$\left(S_{{\mathsf{\ell }}^{\prime }}{S}_{\mathsf{\ell }}\right)\left(X\right)=S_{{\mathsf{\ell }}^{\prime }}\left(P+u(X-P)u\right)=P+u^{\prime }u(X-P)u{u}^{\prime },$$

and $u^{\prime }u=u^{\prime }·u+u^{\prime }\wedge u=\cos \phi -u\wedge u^{\prime }=\cos \phi -\mathfrak{i}\sin \phi =e^{-\mathfrak{i}\phi }$, and $u{u}^{\prime }=e^{\mathfrak{i}\phi }$. Therefore, using that $\mathfrak{i}$ anticommutes with vectors, we have

$$\left(S_{{\mathsf{\ell }}^{\prime }}{S}_{\mathsf{\ell }}\right)\left(X\right)=P+e^{-\mathfrak{i}\phi }(X-P)e^{\mathfrak{i}\phi }=P+(X-P)e^{2\mathfrak{i}\phi }=R_{P,2\phi }\left(X\right).$$

(11)

In particular we note that $S_{\mathsf{\ell }}^2=\mathrm{Id}$.

If the lines ℓ and ${\mathsf{\ell }}^{\prime }$ are parallel, say $\mathsf{\ell }=P+\langle u\rangle$ and ${\mathsf{\ell }}^{\prime }=P^{\prime }+\langle u\rangle$ ($u^2=1$), let $v=P^{\prime }-P$ and $v=v^{\prime }+v^{″}$, with $v^{\prime }\in \langle u\rangle$ and $v^{″}\in {\langle u\rangle }^{\perp }$. Then we have (see Figure 5, Right):

$$\begin{array}{cc}\hfill \left(S_{{\mathsf{\ell }}^{\prime }}{S}_{\mathsf{\ell }}\right)\left(X\right)& =S_{{\mathsf{\ell }}^{\prime }}(P+u(X-P)u)\hfill \\ \hfill & =P^{\prime }+u\left(P-P^{\prime }+u(X-P)u\right)u\hfill \\ \hfill & =P^{\prime }+u(P-P^{\prime })u+X-P\hfill \\ \hfill & =X+v-uvu\hfill \\ \hfill & =X+v^{\prime }+v^{″}-(v^{\prime }-v^{″})\hfill \\ \hfill & =X+2v^{″}.\hfill \end{array}$$

This means that the composition is the translation $T_{2v^{″}}:X\mapsto X+2v^{″}$.

• ⋄11 (Groups of isometries—for future reference). We let $U=\{u_{\phi }=e^{\mathfrak{i}\phi }\phi \in [0,2\pi )\}$, the group of unit (geometric) complex numbers, and $\mathcal{T}=\{T_{v}v\in \mathcal{E}\}$ the group of translations (actually, the map $\mathcal{E}\to \mathcal{T}$, $v\mapsto T_v$ is an isomorphism, as $T_{v+v^{\prime }}=T_v{T}_{v^{\prime }}$ and $T_v=\mathrm{Id}$ if and only if $v=0$).

Fix a point O (call it origin) and consider the set $G_O=\left\{R_{O,\phi }\right|\phi \in [0,2\pi )\}$ of rotations about O. This set is actually a group under the composition operation, as $\mathrm{Id}=R_{O,0}\in G_O$ and

$$R_{O,\phi }{R}_{O,{\phi }^{\prime }}=R_{O,\phi +{\phi }^{\prime }},R^{-1}(O,\phi )=R_{O,-\phi }.$$

Moreover, the map $U\to G_O$, $u_{\phi }\mapsto R_{O,\phi }$, is a group isomorphism.

Define $G=\left\{T_v{R}_{O,\phi }\right|v\in \mathcal{E},\phi \in [0,2\pi )\}$. It is clear that this set contains $\mathcal{T}$. It also contains all rotations $R_{P,\phi }$, for any P and any $\phi$, because $R_{P,\phi }=T_v{R}_{O,\phi }$, where $v=(P-O)(1-u_{\phi })$ (use ⋄6 or check it directly). Now ⋄6 implies that G is a group under the composition operation. Moreover, the map $f_O:\mathcal{E}\times U\to G$, $(v,u_{\phi })\mapsto T_v{R}_{O,\phi }$ is bijective, but it is not a group isomorphism because

$$\begin{array}{cc}\hfill \left(T_v{R}_{O,\phi }\right)\left(T_{v^{\prime }}{R}_{O,{\phi }^{\prime }}\right)& =T_v\left(R_{O,\phi }{T}_{v^{\prime }}\right)R_{O,{\phi }^{\prime }}\hfill \\ \hfill & =T_v{R}_{P,\phi }{R}_{O,{\phi }^{\prime }}[P=O+v^{\prime }{u}_{\phi }{(1-u_{\phi })}^{-1}]\hfill \\ \hfill & =T_v{T}_w{R}_{O,\phi }{R}_{O,{\phi }^{\prime }}[w=(P-O)(1-u_{\phi })=+v^{\prime }{u}_{\phi }]\hfill \\ \hfill & =T_{v+v^{\prime }{u}_{\phi }}{R}_{O,\phi +{\phi }^{\prime }}.\hfill \end{array}$$

This suggests defining in $\mathcal{E}\times U$ the operation $(v,u_{\phi })(v^{\prime },u_{{\phi }^{\prime }})=(v+v^{\prime }{u}_{\phi },u_{\phi }{u}_{{\phi }^{\prime }})$. With this operation, $\mathcal{E}\times U$ is a group, which here we denote by $P=\mathcal{E}⋉U$ (usually called the semi-direct product of $\mathcal{E}$ by U). With this definition, the computation above shows that the map $f_O:P\to G$ is a group isomorphism.

The group P can also be described by using the set of matrices

$$M=\left\{\left(\begin{array}{cc}1& 0\\ v& u_{\phi }\end{array}\right) | v\in \mathcal{E},u_{\phi }\in U\right\},$$

endowed with the product

$$\left(\begin{array}{cc}1& 0\\ v& u_{\phi }\end{array}\right)\left(\begin{array}{cc}1& 0\\ v^{\prime }& u_{{\phi }^{\prime }}\end{array}\right)=\left(\begin{array}{cc}1& 0\\ v+v^{\prime }{u}_{\phi }& u_{\phi }{u}_{{\phi }^{\prime }}\end{array}\right),$$

(12)

as the map $P\to M$, $(v,u_{\phi })\mapsto \left(\begin{array}{cc}1& 0\\ v& u_{\phi }\end{array}\right)$, is manifestly an isomorphism. We note that the product (12) is a slight variation of the matrix product, as instead of $u_{\phi }{v}^{\prime }$ we use $v^{\prime }{u}_{\phi }$. Nevertheless, the product is associative, for this property is easily seen to be equivalent to the commutativity of U. It is also clear that $I_2\in M$ is the neutral element and that

$${\left(\begin{array}{cc}1& 0\\ v& u_{\phi }\end{array}\right)}^{-1}=\left(\begin{array}{cc}1& 0\\ -v{u}_{-\phi }& u_{-\phi }\end{array}\right).$$

A final remark is that under the isomorphisms $M\simeq P\simeq G$ the subgroups of P and M corresponding to $\mathcal{T}$ are $\left\{\right(v,1\left)\right|v\in \mathcal{E}\}$ and $\left\{\left(\begin{array}{cc}1& 0\\ v& 1\end{array}\right)|v\in \mathcal{E}\right\}$, respectively, but we will use the same symbol $\mathcal{T}$ to denote them.

The map $\delta :M\to U$, $\left(\begin{array}{cc}1& 0\\ v& u_{\phi }\end{array}\right)\mapsto \det \left(\begin{array}{cc}1& 0\\ v& u_{\phi }\end{array}\right)=u_{\phi }$ is a surjective homomorphism and its kernel is $\mathcal{T}$. Therefore, $\mathcal{T}$ is a normal subgroup. The corresponding surjective homomorphisms $P\to U$ and $G\to U$ will also be denoted by $\delta$.

Let us describe the action of G on $\mathcal{E}$ in terms of M. The action of $f=T_v{R}_{O,\phi }$ on $\mathcal{E}$ is given by the relation $f\left(X\right)=v+(X-O)u_{\phi }$, which is the second component of the product

$$\left(\begin{array}{cc}1& 0\\ v& u_{\phi }\end{array}\right)\left(\begin{array}{c}1\\ X-O\end{array}\right).$$

This suggests representing points X as vectors $\left(\begin{array}{c}1\\ \mathit{x}\end{array}\right)$, where $\mathit{x}=X-O$, and in this way the action is given by matrix multiplication.

Remark 3. 

Although we are going to need only the group G, it is interesting to point out the following facts. Let S be the set of all gliding symmetries of $\mathcal{E}$ and $\overline{G}=G\bigsqcup S$. Using the results of ⋄6–⋄10 it turns out that $\overline{G}$ is closed under the operation of composition and consequently it is a group. Moreover, it can be seen that $\overline{G}$ is the group of all isometries of $\mathcal{E}$, and that G is the group of all proper isometries, that is, isometries that preserve orientation. The isometries in S reverse the orientation and are called improper isometries.

4. Symmetry Insights About Morley’s Construction

With the notations used in Morley’s construction (Figure 1), we set:

$$R_A=R_{A,2\alpha },R_B=R_{B,2\beta },R_C=R_{C,2\gamma }.$$

(13)

Theorem 2. 

$R_A,R_B,R_C$ have the following properties:

(1) 

$R_A=S_{AZ}{S}_{AB}=S_{AC}{S}_{AY}$

$R_B=S_{BX}{S}_{BC}=S_{BA}{S}_{BZ}$

$R_C=S_{CY}{S}_{CA}=S_{CB}{S}_{CX}$

(2) 

$R_B{R}_C=R_{X,2(\pi -{\alpha }^{\prime })}=R_{X,2({\displaystyle \frac{\pi }{3}}-\alpha )}$

$R_C{R}_A=R_{Y,2(\pi -{\beta }^{\prime })}=R_{X,2({\displaystyle \frac{\pi }{3}}-\beta )}$

$R_A{R}_B=R_{Z,2(\pi -{\gamma }^{\prime })}=R_{Z,2({\displaystyle \frac{\pi }{3}}-\gamma )}$

(3) 

$R_A^3=S_{AC}{S}_{AB}$, $R_B^3=S_{BA}{S}_{BC}$, $R_C^3=S_{CB}{S}_{CA}$.

(4) 

$R_A^3{R}_B^3{R}_C^3=\mathrm{Id}$.

Proof. 

Because of the cyclic symmetry, it is enough to prove one relation in each case.

(1)

Follows directly from Equation (11): $S_{AZ}{S}_{AB}=R_{A,2\phi (AB,AZ)}=R_{A,2\alpha }=R_A$ and likewise $S_{AC}{S}_{AY}=R_{A,2\phi (AY,AC)}=R_{A,2\alpha }=R_A$.

(2)

By (1), we can write $R_B{R}_C=S_{BX}{S}_{BC}{S}_{CB}{S}_{CX}=S_{XB}{S}_{XC}$, which by Equation (11) is equal to $R_{X,2\phi (XC,XB)}=R_{X,2(2\pi -{\alpha }^{\prime })}=R_{X,2(\pi -{\alpha }^{\prime })}=R_{X,2({\displaystyle \frac{\pi }{3}}-\alpha )}$ (in the second step the summand $2\pi$ can be omitted as its contribution to the rotation is null).

(3)

We have $R_A^3=R_{A,6\alpha }$, and this is equal to $S_{AC}{S}_{AB}$ by Equation (11).

(4)

By (3), $R_A^3{R}_B^3{R}_C^3=S_{AC}{S}_{AB}{S}_{BA}{S}_{BC}{S}_{CB}{S}_{CA}$, which is the identity because $S_{AB}{S}_{BA}$, $S_{BC}{S}_{CB}$, $S_{AC}{S}_{CA}$ are the identity.

□

In what follows, the unit complex number $e^{\mathfrak{i}\phi }$ will be denoted by $u_{\phi }$. In particular, we have $R_{P,\phi }\left(X\right)=P+(X-P)u_{\phi }$. Moreover, $R_{P,u_{\phi }}$ will be used as an alias for $R_{P,\phi }$.

Theorem 3 

(Fixed point of the composition of two rotations). If $u_{\phi }{u}_{\psi }\ne 1$, the composition $R=R_{Q,\psi }{R}_{P,\phi }$ has a unique fixed point F, which is given by the expression

$$F=P+(Q-P){\displaystyle \frac{1-u_{\psi }}{1-u_{\phi }{u}_{\psi }}},$$

(14)

and in this case $R=R_{F,\phi +\psi }$. Otherwise, $R_{Q,\psi }{R}_{P,\phi }=T_v$, where $v=(Q-P)(1-u_{\psi })$.

Proof. 

For any point X, its transform $X^{\prime }$ is

$$\begin{array}{cc}\hfill X^{\prime }& =R_{Q,\psi }\left(R_{P,\phi }\left(X\right)\right)\hfill \\ \hfill & =R_{Q,\psi }(P+(X-P)u_{\phi })\hfill \\ \hfill & =Q+\left(P-Q+(X-P)u_{\phi }\right)u_{\psi }\hfill \\ \hfill & =Q-(Q-P)u_{\psi }+(X-P)u_{\phi }{u}_{\psi }.\hfill \end{array}$$

If $u_{\phi }{u}_{\psi }=1$, this equation says that $X^{\prime }=X+(Q-P)(1-u_{\psi })=X+v$, or $X^{\prime }=T_v\left(X\right)$. So we may assume that $u_{\phi }{u}_{\psi }\ne 1$. If F is fixed by $R=R_{Q,\psi }{R}_{P,\phi }$, then $F=P+(F-P)$ is equal to $F^{\prime }=Q-(Q-P)u_{\psi }+(F-P)u_{\phi }{u}_{\psi }$; hence,

$$\begin{array}{cc}\hfill & Q-(Q-P)u_{\psi }+(F-P)u_{\phi }{u}_{\psi }=P+(F-P)\hfill \\ \hfill \iff & (Q-P)(1-u_{\psi })=(F-P)(1-u_{\phi }{u}_{\psi }).\hfill \end{array}$$

and solving for $F-P$ we get a unique F given by the claimed formula.

$R=R_{F,\phi +\psi }$ is established by calculating $R_{F,\phi +\psi }\left(X\right)=F+(X-F)u_{\phi }{u}_{\psi }$ using the expression for F given by Equation (14):

$$\begin{array}{cc}\hfill F+(X-F)u_{\phi }{u}_{\psi }& =P+(Q-P)\frac{1-u_{\psi }}{1-u_{\phi }{u}_{\psi }}+(X-P-(Q-P)\frac{1-u_{\psi }}{1-u_{\phi }{u}_{\psi }})u_{\phi }{u}_{\psi }\hfill \\ \hfill & =P+(Q-P)(1-u_{\psi })+(X-P)u_{\phi }{u}_{\psi }\hfill \\ \hfill & =Q+(P-Q)-(P-Q)(1-u_{\psi })+(X-P)u_{\phi }{u}_{\psi }\hfill \\ \hfill & =Q+(Q-P)u_{\psi }+(X-P)u_{\phi }{u}_{\psi },\hfill \end{array}$$

which agrees with $R\left(X\right)$. □

Corollary 1 

(Determination of $X,Y,Z$). Let $u_A=e^{2\mathfrak{i}\alpha }$, $u_B=e^{2\mathfrak{i}\beta }$, $u_C=e^{2\mathfrak{i}\gamma }$. Then we have:

$$\begin{array}{cc}\hfill X& =C+(B-C)\frac{1-u_B}{1-u_B{u}_C}=C+(B-C)\frac{\sin \beta }{\sin (\beta +\gamma )}{e}^{-\mathfrak{i}\gamma }=B+(C-B)\frac{\sin \gamma }{\sin (\beta +\gamma )}{e}^{\mathfrak{i}\beta },\hfill \\ \hfill Y& =A+(C-A)\frac{1-u_C}{1-u_C{u}_A}=A+(C-A)\frac{\sin \gamma }{\sin (\gamma +\alpha )}{e}^{-\mathfrak{i}\alpha }=C+(A-C)\frac{\sin \alpha }{\sin (\gamma +\alpha )}{e}^{\mathfrak{i}\gamma }\hfill \\ \hfill Z& =B+(A-B)\frac{1-u_A}{1-u_A{u}_B}=B+(A-B)\frac{\sin \alpha }{\sin (\alpha +\beta )}{e}^{-\mathfrak{i}\beta }=A+(B-A)\frac{\sin \beta }{\sin (\alpha +\beta )}{e}^{\mathfrak{i}\alpha }.\hfill \end{array}$$

We only need to work out the expressions for X, as those for Y and Z follow from the cyclic arrangement. The first expression is obtained by applying the formula of Theorem 3 to the composition $R_B{R}_C$, as X is its fixed point. For the second expression of X note that

$$\begin{array}{cc}\hfill 1-u_B& =1-e^{2\mathfrak{i}\beta }=1-\cos 2\beta -\mathfrak{i}\sin 2\beta =2{\sin }^2\beta -2\mathfrak{i}\sin \beta \cos \beta \hfill \\ \hfill & =-2\mathfrak{i}\sin \beta (\cos \beta +\mathfrak{i}\sin \beta )=-2\mathfrak{i}\sin \beta e^{\mathit{i}\beta },\hfill \end{array}$$

and hence $\frac{1-u_B}{1-u_B{u}_C}=\frac{\sin \beta e^{\mathfrak{i}\beta }}{\sin (\beta +\gamma )e^{\mathfrak{i}(\beta +\gamma )}}=\frac{\sin \beta }{\sin (\beta +\gamma )}{e}^{-\mathfrak{i}\gamma }$. Finally, the third expression for X is derived from the second as follows:

$$\begin{array}{cc}\hfill C+(B-C)\frac{\sin \beta }{\sin (\beta +\gamma )}{e}^{-\mathfrak{i}\gamma }& =B+(C-B)-(C-B)\frac{\sin \beta }{\sin (\beta +\gamma )}{e}^{-\mathfrak{i}\gamma }\hfill \\ \hfill & =B+(C-B)(1-\frac{\sin \beta }{\sin (\beta +\gamma )}{e}^{-\mathfrak{i}\gamma })\mathrm{and}\hfill \\ \hfill \sin (\beta +\gamma )-\sin \beta e^{-\mathfrak{i}\gamma }& =\sin \beta \cos \gamma +\cos \beta \sin \gamma -\sin \beta \cos \gamma +\mathfrak{i}\sin \beta \sin \gamma \hfill \\ \hfill & =\sin \gamma (\cos \beta +\mathfrak{i}\sin \beta )=\sin \gamma e^{\mathfrak{i}\beta }.\square \hfill \end{array}$$

The formulas in Corollary 1 can be used to draw Morley’s triangle and then the trisectors of the three angles, as shown in Figure 6.

Remark 4. 

The second and third expressions for X can be inferred from the law of sines, which has not been used to obtain them. Indeed, with the notations of Figure 1, $X-C=\frac{XC}{BC}(B-C)e^{-\mathfrak{i}\gamma }$, where $\frac{XC}{BC}=\frac{\sin \beta }{\sin {\alpha }^{\prime }}$ by the sine theorem, and $\sin {\alpha }^{\prime }=\sin (\beta +\gamma )$. Similarly, $X-B=\frac{BX}{BC}(C-B)e^{\mathfrak{i}\beta }=\frac{\sin \gamma }{\sin {\alpha }^{\prime }}(C-B)e^{\mathfrak{i}\beta }=(B-A)\frac{\sin \gamma }{\sin (\beta +\gamma )}{e}^{\mathfrak{i}\beta }$. Anyway, the law of sines can be obtained easily using the Wessel algebra (see ⋄19).

• ⋄12 (Criteria for a triangle to be equilateral). Let $ABC$ be a triangle in ${\mathcal{E}}_2$ and O a given point (origin). Let ${\omega }_0,{\omega }_1,{\omega }_2\in \mathbb{C}$. Then the expression

  $$(A-O){\omega }_0+(B-O){\omega }_1+(C-O){\omega }_2$$

  (15)

  is independent of O if and only if ${\omega }_0+{\omega }_1+{\omega }_2=0$. This follows immediately from the following computation, where $O^{\prime }$ is any other point:

  $$\begin{array}{cc}\hfill & (A-O){\omega }_0+(B-O){\omega }_1+(C-O){\omega }_2\hfill \\ \hfill & =(A-O^{\prime }){\omega }_0+(O^{\prime }-O){\omega }_0+(B-O^{\prime }){\omega }_1+(O^{\prime }-O){\omega }_1+(C-O^{\prime }){\omega }_2+(O^{\prime }-O){\omega }_2\hfill \\ \hfill & =(A-O^{\prime }){\omega }_0+(B-O^{\prime }){\omega }_1+(C-O^{\prime }){\omega }_2+(O^{\prime }-O)({\omega }_0+{\omega }_1+{\omega }_2).\hfill \end{array}$$

If we now apply this result using ${\omega }_0=1,{\omega }_1=\omega ,{\omega }_2={\omega }^2$, where $\omega =e^{{\displaystyle \frac{2\pi }{3}}\mathfrak{i}}$ (note that ${\omega }^3=1$ and hence $1+\omega +{\omega }^2=({\omega }^3-1)/(\omega -1)=0$), then in particular we have

$$(A-O)+(B-O)\omega +(C-O){\omega }^2=(A-G)+(B-G)\omega +(C-G){\omega }^2$$

where G is the barycenter of $ABC$. With this we can prove that $ABC$ is equilateral if and only if $(A-O)+(B-O)\omega +(C-O){\omega }^2=0$, that is, if and only if $\mathit{x}+\mathit{y}\omega +\mathit{z}{\omega }^2=0$, where $\mathit{x}=A-G$, $\mathit{y}=B-G$, $\mathit{z}=C-G$. If the triangle is equilateral, then $\mathit{y}=\mathit{x}\omega$ and $\mathit{z}=\mathit{y}\omega =\mathit{x}{\omega }^2$, and hence $\mathit{x}+\mathit{y}\omega +\mathit{z}{\omega }^2=\mathit{x}+\mathit{x}{\omega }^2+\mathit{x}{\omega }^4=\mathit{x}(1+{\omega }^2+\omega )=0$. Conversely, assume that $\mathit{x}+\mathit{y}\omega +\mathit{z}{\omega }^2=0$. This, together with $\mathit{x}+\mathit{y}+\mathit{z}=0$, implies that $\mathit{y}(\omega -1)+\mathit{z}({\omega }^2-1)=0$, or $\mathit{y}+\mathit{z}(\omega +1)=0$, or $\mathit{y}=\mathit{z}{\omega }^2$ (because $\omega +1=-{\omega }^2$), or $\mathit{z}=\mathit{y}\omega$. Hence, we also have $\mathit{x}=-(\mathit{y}+\mathit{z})=-\mathit{y}(1+\omega )=\mathit{y}{\omega }^2$, or $\mathit{y}=\mathit{x}\omega$, and these facts imply that $ABC$ is equilateral.

The special case $O=A$ tells us that $ABC$ is equilateral if and only if

$$(B-A)\omega +(C-A){\omega }^2=0,\mathrm{or}B-A=(A-C)\omega .$$

(16)

5. Proof of Theorem 1 with Wessel’s Algebra

• ⋄13 (Notations and conventions). Let us set (cf. Figure 1):

  $$\mathit{a}=C-B,\mathit{b}=A-C,\mathit{c}=B-A,$$

  so that $\mathit{a}=a{\mathit{a}}^{\star }$, $\mathit{b}=b{\mathit{b}}^{\star }$, $\mathit{c}=c{\mathit{c}}^{\star }$ (${\mathit{a}}^{\star },{\mathit{b}}^{\star },{\mathit{c}}^{\star }$ unit vectors). Then we have the following relations:

  $${\mathit{b}}^{\star }=-{\mathit{a}}^{\star }{e}^{-\mathfrak{i}3\gamma },{\mathit{c}}^{\star }=-{\mathit{b}}^{\star }{e}^{-\mathfrak{i}3\alpha },{\mathit{a}}^{\star }=-{\mathit{c}}^{\star }{e}^{-\mathfrak{i}3\beta }.$$

  (17)

  Notice, for example, that ${\mathit{b}}^{\star }={\mathit{a}}^{\star }{e}^{\mathfrak{i}(\pi -3\gamma )}=-{\mathit{a}}^{\star }{e}^{-\mathfrak{i}3\gamma }$.
• ⋄14 (Proof). It is based on the expressions for $X,Y,Z$ obtained in Corollary 1. With the notations and conventions declared in

  Table 3. For a given angle $\theta$, the variables ${\theta }^{\prime }$, ${\theta }^{″}$, ${\theta }^{\ast }$, ${\theta }^{\ast \ast }$ are defined in terms of $\theta$. Since clearly ${\theta }^{\prime }+{\theta }^{\ast \ast }={\theta }^{″}+{\theta }^{\ast }=\pi$, then $\sin {\theta }^{\ast \ast }=\sin {\theta }^{\prime }$, $\cos {\theta }^{\ast \ast }=-\cos {\theta }^{\prime }$, $\sin {\theta }^{\ast }=\sin {\theta }^{″}$, $\cos {\theta }^{\ast }=-\cos {\theta }^{″}$. Note that according to the convention declared in Figure 1 the value ${\theta }^{″}$ agrees with $\overline{\theta }$. We also have that $\sin (\beta +\gamma )=\sin (\frac{\pi }{3}-\alpha )=\sin {\alpha }^{\ast }$ and similarly $\sin (\gamma +\alpha )=\sin {\beta }^{\ast }$, $\sin (\alpha +\beta )=\sin {\gamma }^{\ast }$.

  ${\theta }^{\prime }$	${\theta }^{″}$	${\theta }^{\ast }$	${\theta }^{\ast \ast }$
  $\frac{\pi }{3}+\theta$	$\frac{2\pi }{3}+\theta$	$\frac{\pi }{3}-\theta$	$\frac{2\pi }{3}-\theta$

  , the third expression of Z and the second of Y in Corollary 1 take the form

  $$Z=A+{\mathit{c}}^{\star }c\frac{\sin \beta }{\sin {\gamma }^{\ast }}{e}^{\mathfrak{i}\alpha },Y=A-{\mathit{b}}^{\star }b\frac{\sin \gamma }{\sin {\beta }^{\ast }}{e}^{-\mathfrak{i}\alpha }.$$

  Since ${\mathit{c}}^{\star }=-{\mathit{b}}^{\star }{e}^{-\mathfrak{i}3\alpha }$, by (17), we readily get

  $$Z-Y={\mathit{b}}^{\star }\left(\frac{b\sin \gamma }{\sin {\beta }^{\ast }}-\frac{c\sin \beta }{\sin {\gamma }^{\ast }}{e}^{-\mathfrak{i}\alpha }\right)e^{-\mathfrak{i}\alpha }.$$

  Now use Equation (A2) to replace b and c, and then the identity (A4) to obtain

  $$\begin{array}{cc}\hfill Z-Y& =4\delta \sin \beta \sin \gamma {\mathit{b}}^{\star }(\sin {\beta }^{\ast \ast }-\sin {\gamma }^{\ast \ast }{e}^{-\mathfrak{i}\alpha })e^{-\mathfrak{i}\alpha }.\hfill \end{array}$$

  It can be immediately seen that ${\beta }^{\ast \ast }={\gamma }^{\prime }+\alpha$, and hence $\sin {\beta }^{\ast \ast }=\sin {\gamma }^{\prime }\cos \alpha +\cos {\gamma }^{\prime }\sin \alpha$. On the other hand, $\sin {\gamma }^{\ast \ast }{e}^{-\mathfrak{i}\alpha }=\sin {\gamma }^{\prime }\cos \alpha -\mathfrak{i}\sin {\gamma }^{\prime }\sin \alpha$. Therefore,

  $$\sin {\beta }^{\ast \ast }-\sin {\gamma }^{\ast \ast }{e}^{-\mathfrak{i}\alpha }=\cos {\gamma }^{\prime }\sin \alpha +\mathfrak{i}\sin {\gamma }^{\prime }\sin \alpha =\sin \alpha e^{\mathfrak{i}{\gamma }^{\prime }}.$$

  Since $e^{\mathfrak{i}{\gamma }^{\prime }}{e}^{-\mathfrak{i}\alpha }=e^{\mathfrak{i}({\gamma }^{\prime }-\alpha )}=e^{\mathfrak{i}(\beta +2\gamma )}$, we get

  $$Z-Y=4\delta \sin \alpha \sin \beta \sin \gamma {\mathit{b}}^{\star }{e}^{\mathfrak{i}(\beta +2\gamma )}.$$

  (18)

  The expressions for $X-Z$ and $Y-X$ are obtained from (18) by cyclic permutations:

  $$X-Z=4\delta \sin \alpha \sin \beta \sin \gamma {\mathit{c}}^{\star }{e}^{\mathfrak{i}(\gamma +2\alpha )},$$

  (18a)

  $$Y-X=4\delta \sin \alpha \sin \beta \sin \gamma {\mathit{a}}^{\star }{e}^{\mathfrak{i}(\alpha +2\beta )}.$$

  (18b)

  In particular $XYZ$ is equilateral: $XY=YZ=ZX=4\delta \sin \alpha \sin \beta \sin \gamma$.

Remark 5. 

The expressions (18) tell us not only the length of the sides of Morley’s triangle $XYZ$, but also the orientation of its sides with respect to the triangle $ABC$.