Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 111 / 4

Reversing a word for $w$ gives a word of the same length for $w^{-1}$ because every generator is an involution. Applying the same argument to $w^{-1}$ provesA shortest word for $w$ followed by $s_i$ gives ${\ell }_S(w{s}_i)\le {\ell }_S(w)+1$. Conversely, $w=(w{s}_i)s_i$ gives ${\ell }_S(w)\le {\ell }_S(w{s}_i)+1$. Hence

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We prove the assertion by induction on $n$. Write $w=av$, where $a=s_{i_1}$ and $v=s_{i_2}\cdots s_{i_n}$; both displayed words are reduced. The first clause of the folding condition and part a imply that $\ell (vs)$ is either $n-2$ or $n$.

If $\ell (vs)=n-2$, the induction hypothesis deletes one unique letter from the reduced word for $v$. Prefixing $a$ gives the required deletion from $w$. If another deletion were possible, induction rules out another internal position, while deletion of $a$ would give $ws=v$ and hence $vs=av$, contradicting the lengths $n-2$ and $n$.

If $\ell (vs)=n$, both $av$ and $vs$ increase the length of $v$. Since $\ell (avs)=\ell (ws)=n-1$, the other alternative in the folding condition must hold:Thus $ws=avs=v$, which deletes the first letter. An additional internal deletion would give $v=a{v}^{\prime }$ for a word $v^{\prime }$ of length $n-2$; multiplying by $a$ would make the length-$n$ element $av=w$ equal to $v^{\prime }$, impossible. The deletion position is therefore unique in every case.

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If $m_{ij}<\infty$, the braid relation in a Coxeter group isFor $m_{ij}=2$ this is commutation; no relation is imposed when $m_{ij}=\infty$. Two reduced expressions are braid equivalent when one can be transformed into the other by finitely many replacements of one side of such a relation by the other.

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First suppose $(W,S)$ is a Coxeter system. The usual exchange condition implies that multiplication by a simple generator changes length by exactly one. Let $w=t_1\cdots t_n$ be reduced and suppose both $s_{i}w$ and $w{s}_j$ have length $n+1$. The length of $s_{i}w{s}_j$ is therefore either $n+2$ or $n$. In the latter case, apply exchange to the reduced word $s_i{t}_1\cdots t_n$ followed by $s_j$. If exchange deleted one of the $t_k$, multiplying the resulting equality on the left by $s_i$ would express the length-$n+1$ element $w{s}_j$ using only $n-1$ generators. Hence exchange must delete the initial $s_i$, giving $s_{i}w{s}_j=w$. This is exactly the folding condition.

Conversely, suppose the folding condition holds, and let $W_M$ be the abstract Coxeter group with generators $S$ and matrix $(m_{ij})$. The defining relations hold in $W$, so there is a surjective homomorphismIt remains to prove injectivity. Take any word in the kernel. If its image word in $W$ is not reduced, choose its shortest nonreduced prefix $ut$, where $u$ is reduced and $t$ is its last generator. Part b gives $ut=v$, where $v$ is obtained by deleting one letter from $u$. Hence $u=vt$, and $u$ and $vt$ are two reduced expressions for the same element. By the assumed braid-equivalence theorem they are related by braid moves. Those moves are defining relations in $W_M$, after which the end of the prefix becomes $vtt$ and $t^2=1$ shortens the original word by two.

Repeating this process turns the kernel word, using only Coxeter relations, into a word that is reduced in $W$. Since its image is the identity, that reduced word is empty. The original word is therefore already the identity in $W_M$, so $\ker \pi =1$. Hence $W\cong W_M$ is the Coxeter group with generators $S$.

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