# Morley’s Theorem, Alain Connes’s Proof

At the end of 19th century, the British mathematician Frank Morley discovered the following surprising property :

The three points of intersection of the adjacent trisectors of the angles of any triangle are the vertices of an equilateral triangle.

There are many proofs of this result, but one of them given by Alain Connes is especially beautiful and interesting as a group theoretic property of the action of the affine group on the line. We propose an overview of his proof.

AFFINE GROUP

Let $k$ be a commutative field and $G$ the affine group over $k$ that is to say the group of mappings ${g_{a,b}\left(x\right)=ax+b}$ where ${a\neq0}$ and $x \in k$. In fact, $G$ can be represented by the group of 2×2  invertible matrices of the form

$\displaystyle \left(\begin{array}{cc} a & b\\ 0 & 1\end{array}\right)\textrm{ where }a\in k^{*},b\in k$

working as follows

$\displaystyle \left(\begin{array}{cc} a & b\\ 0 & 1\end{array}\right)\left(\begin{array}{c} x\\ 1\end{array}\right)=\left(\begin{array}{c} ax+b\\ 1\end{array}\right)$

Next, we define a group homomorphism ${\delta:G\rightarrow k^{*}}$ where ${k^{*}}$ is the multiplicative group of nonzero elements of k

$\displaystyle \delta\left(g_{a,b}\right)=a$

Clearly, ${\ker\delta}$ is the group of translations of k. We denote the translational part by

$\displaystyle \tau\left(g_{a,b}\right)=b$

Finally, when ${a\neq1}$ there is a unique fixed point ${x_{0}=ax_{0}+b}$ or ${x_{0}=b/\left(1-a\right)}$ that we denote by

$\displaystyle \mathrm{fix}\left(g_{a,b}\right)=\frac{b}{1-a}$

CONNES’S THEOREM

The proof of the Morley’s theorem will appear as a direct consequence of the following theorem. Although it could seem rather abstract at first look, we will see later that it is really the solution of the problem.

Connes’s Theorem : Let ${f,g,h\in G}$ be such that ${fg}$, ${gh}$, ${hf}$ and ${fgh}$ are not translations and let ${j=\delta\left(fgh\right)}$. The two following conditions are equivalent

(1) ${f^{3}g^{3}h^{3}=1}$

(2) ${j^{3}=1}$ and ${\alpha+j\beta+j^{2}\gamma=0}$ where ${\alpha=\mathrm{fix}\left(fg\right)}$, ${\beta=\mathrm{fix}\left(gh\right)}$, ${\gamma=\mathrm{fix}\left(hf\right)}$

To prove this equivalence, we notice that

$\displaystyle f^{3}g^{3}h^{3}=1\textrm{ is equivalent to }\delta\left(f^{3}g^{3}h^{3}\right)=1\textrm{ and }\tau\left(f^{3}g^{3}h^{3}\right)=0$

The outline of the proof can be summarized as follows

With the help of a computer algebra system (like the open source software Maxima) we compute with no effort

$\displaystyle f^{3}g^{3}h^{3}=\left(\begin{array}{cc} A & B\\ 0 & 1\end{array}\right)$

with

$\displaystyle A=a_{f}^{3}a_{g}^{3}a_{h}^{3}$

$\displaystyle B=a_{f}^{3}a_{g}^{3}a_{h}^{2}b_{h}+a_{f}^{3}a_{g}^{3}a_{h}b_{h}+a_{f}^{3}a_{g}^{3}b_{h}+a_{f}^{3}a_{g}^{2}b_{g}+a_{f}^{3}a_{g}b_{g}+a_{f}^{3}b_{g}+a_{f}^{2}b_{f}+a_{f}b_{f}+b_{f}$

where $a_{f}, b_{f}, a_{g}, b_{g}, a_{h}, b_{h}$ are the “a and b” coefficients of $f$, $g$, $h$ respectively.

First, we prove that ${\delta\left(f^{3}g^{3}h^{3}\right)=1}$ is the same as ${j^{3}=1}$. For this, we simply check that ${\delta\left(f^{3}g^{3}h^{3}\right)=j^{3}}$ (the equality before the last uses the homomorphism property of ${\delta}$)

$\displaystyle \delta\left(f^{3}g^{3}h^{3}\right)=A=a_{f}^{3}a_{g}^{3}a_{h}^{3}=\left(a_{f}a_{g}a_{h}\right)^{3}=\left(\delta f\cdot\delta g\cdot\delta h\right)^{3}=\left[\delta\left(fgh\right)\right]^{3}=j^{3}$

Second, we prove that ${\tau\left(f^{3}g^{3}h^{3}\right)=0}$ is the same as ${\alpha+j\beta+j^{2}\gamma=0}$. For this, we notice that ${\tau\left(f^{3}g^{3}h^{3}\right)=B}$. We can simplify B as follows

$\displaystyle B=\left(a_{f}a_{g}\right)^{3}\left(a_{h}^{2}+a_{h}+1\right)b_{h}+a_{f}^{3}\left(a_{g}^{2}+a_{g}+1\right)b_{g}+\left(a_{f}^{2}+a_{f}+1\right)b_{f}$

In his paper, Connes proposes the following factorization

$\displaystyle \tau\left(f^{3}g^{3}h^{3}\right)=-ja_{f}^{2}a_{g}\left(a_{f}-j\right)\left(a_{g}-j\right)\left(a_{h}-j\right)\left(\alpha+j\beta+j^{2}\gamma\right)$

The factor ${\left(a_{f}-j\right)}$ cannot be zero because ${a_{f}-j=a_{f}\left(1-a_{g}a_{h}\right)}$ and ${gh}$ is not a translation (hypothesis of the theorem) which means ${a_{g}a_{h}\neq1}$. The same occurs for the factors ${\left(a_{g}-j\right)}$ and ${\left(a_{h}-j\right)}$. Therefore, ${\tau\left(f^{3}g^{3}h^{3}\right)=0}$ is equivalent to ${\alpha+j\beta+j^{2}\gamma=0}$.

MORLEY’S THEOREM

We take ${k=\mathbb{C}}$ and define $f$, $g$, $h$ as follows

• $f$ is the rotation of center $A$ and angle $2a$ where ${3a=\widehat{BAC}}$
• $g$ is the rotation of center $B$ and angle $2b$ where ${3b=\widehat{CBA}}$
• $h$ is the rotation of center $C$ and angle $2c$ where ${3c=\widehat{ACB}}$

Now, consider the point ${\alpha}$ of intersection of the trisectors of angles $A$ and $B$ closest to the side ${AB}$. The rotation $g$ transforms ${\alpha}$ to ${\alpha'}$ and the rotation $f$ transforms ${\alpha'}$ to ${\alpha}$. Therefore ${\alpha}$ is a fixed point of $f g$. Similarly, ${\beta}$ is a fixed point of $g h$ and ${\gamma}$ is a fixed point of $h f$.

Next, ${f^{3}}$ is a rotation with center $A$ and angle $6a$. It is equivalent to a product of two reflections ${S_{AC}S_{AB}}$. Similarly, ${g^{3}}$ is equivalent to a product of two reflections ${S_{AB}S_{BC}}$ and ${h^{3}}$ is equivalent to a product of two reflections ${S_{BC}S_{AC}}$.

We deduce

$\displaystyle f^{3}g^{3}h^{3}=S_{AC}S_{AB}S_{AB}S_{BC}S_{BC}S_{AC}=1$

Thus, we are in the conditions of the Connes’s theorem, which implies ${j^{3}=1}$ and ${\alpha+j\beta+j^{2}\gamma=0}$. Since $k = \mathbb{C}$, the equality ${j^{3}=1}$ means that ${j}$ is the cube root of unity ${j=e^{i2\pi/3}}$. The relation can be interpreted in the complex plane as follows

$\displaystyle \alpha+j\beta+j^{2}\gamma=0$

$\displaystyle \alpha-\beta-j^{2}\beta+j^{2}\gamma=0$

$\displaystyle \alpha-\beta=-j^{2}\left(\gamma-\beta\right)$

$\displaystyle \alpha-\beta=e^{i\pi/3}\left(\gamma-\beta\right)$

This means that vector ${\beta\alpha}$ is obtained by rotation of angle ${\pi/3}$ from vector ${\beta\gamma}$. The same angle ${\pi/3}$ occurs in the two other cases which proves that the triangle ${\widehat{\alpha\beta\gamma}}$ is equilateral.

REFERENCES

1. A New Proof of Morley’s Theorem, Alain Connes, IHES, 1998, pp. 43-46
2. Newsletter European Mathematical Society, December 2004, Issue 54