Set
$$
a_{ij}=2\cos\left(\frac{\pi}{m_{ij}}\right),
\qquad c_i=x_1\cdots x_i,
\qquad c_0=1.
$$
In the <Geometric representation of a Coxeter group>,
$$
\sigma(x_i)e_j=e_j+a_{ij}e_i,
$$
including $i=j$, since $a_{ii}=2\cos\pi=-2$. As $\beta_i=\sigma(c_{i-1})e_i$, telescoping gives
$$
\sigma(c_i)e_j-\sigma(c_{i-1})e_j
=a_{ij}\beta_i.
$$
Summing from $i=1$ to $j-1$ yields
$$
\beta_j=e_j+\sum_{i<j}a_{ij}\beta_i,
$$
or equivalently
$$
e_j=\beta_j-\sum_{i<j}2\cos\left(\frac{\pi}{m_{ij}}\right)\beta_i.
$$
Summing the same telescoping identity all the way to $n$ and substituting this first formula gives
$$
\begin{aligned}
\sigma(c)e_j
&=e_j+\sum_{i=1}^na_{ij}\beta_i\\
&=\beta_j+\sum_{i\geq j}2\cos\left(\frac{\pi}{m_{ij}}\right)\beta_i,
\end{aligned}
$$
which is the second required identity.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 111 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-111/4/b/solution