Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 111 / 3

Restrict the Coxeter matrix of $(W,S)$ to $T\times T$ and let $W^{\prime }$ be the Coxeter group defined by that matrix. Sending its generators to the corresponding elements of $T$ gives a surjection $W^{\prime }\to W_T$. The Geometric representation of a Coxeter group for $W^{\prime }$ is the restriction of the geometric representation for $W$ to the span of the simple roots indexed by $T$. Faithfulness of the geometric representation makes the map injective. Hence

Part (a) gives a reduced $T$-word for $w$. It is also reduced as an $S$-word, since a shorter $S$-word would contradict the length function obtained from the restricted geometric representation. By Matsumoto theorem, this expression and the given reduced expression $s_1\cdots s_k$ differ by braid moves. Every braid move starting with letters in $T$ replaces them by the same two letters in the opposite alternating order, so it never introduces a generator outside $T$. Therefore every $s_i$ lies in $T$.

By part (a), every full subdiagram gives a standard parabolic Coxeter subgroup. In the $B_{n+1}$ diagram, the $n$ vertices away from the endpoint incident to the edge labelled four form an $A_n$ chain. Thus

The $E_8$ diagram is a trivalent tree whose three arms have lengths $1,2,4$ beyond the central vertex. Removing the endpoint of the arm of length one leaves a chain of seven vertices, hence an $A_7$ subdiagram. Removing the outer endpoint of the arm of length two leaves arms of lengths $1,1,4$, the $D_7$ diagram. Consequently

Realize the generators as affine reflections of ${\mathbb{R}}^2$:The reflecting lines for $s$ and $u$ are perpendicular, while the line $x+y=1$ meets each at angle $\pi /4$. Henceso the presentation maps onto this affine reflection group. Butand $g$ has infinite order. Thus $W$ is infinite. The finite Coxeter group $W(B_4)$ is the signed symmetric group on four letters and has order $2^{4}4!$. An infinite group cannot embed in it, so $W$ is not a subgroup of $W(B_4)$.