For $f\in C$, duality gives
$$
(\sigma^*(w)f)(e_i)=f(\sigma(w^{-1})e_i).
$$
The <positive-root criterion for Coxeter length> says
$$
\ell(x_iw)>\ell(w)
\quad\Longleftrightarrow\quad
\sigma(w^{-1})e_i
$$
is a positive root. Its basis coefficients are nonnegative and not all zero, so $f(\sigma(w^{-1})e_i)>0$ and $\sigma^*(w)(C)\subseteq A_i$. If the length decreases, that root is negative and the same calculation gives $\sigma^*(w)(C)\subseteq A_i^-$.

If $\sigma^*(w)$ is the identity and $w\ne1$, choose a left descent $x_i$ from the first letter of a reduced expression for $w$. The preceding result gives
$$
C=\sigma^*(w)(C)\subseteq A_i^-,
$$
although $C\subseteq A_i$. The two open half-spaces are disjoint, a contradiction. Thus the <Dual geometric representation of a Coxeter group> is faithful.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 111 / 3 / b / Solution

{c}
{title2=$\sigma^*$}

The dual geometric representation acts on $V^*$ by
$$
(\sigma^*(w)f)(v)=f(\sigma(w^{-1})v).
$$
The open fundamental chamber consists of the functionals positive on every simple basis vector.


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 Incoming links: Dual geometric representation of a Coxeter group