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

The Coxeter number is the order of a Coxeter element. In the stated families the values areHere $D_{2n}$ is the dihedral group of order $2n$, $S_n$ is the symmetric group, $S{S}_n$ is the signed symmetric group, and $ES{S}_n$ is the even signed symmetric group.

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SetIn the Geometric representation of a Coxeter group,including $i=j$, since $a_{ii}=2\cos \pi =-2$. As ${\beta }_i=\sigma (c_{i-1})e_i$, telescoping givesSumming from $i=1$ to $j-1$ yieldsor equivalentlySumming the same telescoping identity all the way to $n$ and substituting this first formula giveswhich is the second required identity.

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Let $B$ be the matrix whose $i$th column consists of the coordinates of ${\beta }_i$ in the basis $(e_j)$. Define the upper-triangular matrix $U$ and lower-triangular matrix $L$ byThe first identity in part b says $I=BU$, so $B=U^{-1}$. The second says that the matrix $C$ of $\sigma (c)$ is $C=BL=U^{-1}L$. Since $U$ has diagonal entries one, $\det U=1$, and thereforeThus $\det (tU-L)$ is the characteristic polynomial of the Coxeter element in its geometric representation.

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At $t=1$, the diagonal entries of $U-L$ are $2$, while every off-diagonal $(i,j)$ entry isHence $U-L$ is exactly the Coxeter Gram matrix $G(W)$, and part c givesFor a Finite Coxeter group the Gram matrix is positive definite, and for a Hyperbolic Coxeter group it is nondegenerate with Lorentzian signature. In either case $\det (I-C)\ne 0$, so $1$ is not an eigenvalue of $C$ and the Coxeter element fixes no nonzero vector.

For an Affine Coxeter group, the Gram form has a nonzero radical. If $0\ne v\in \mathrm{rad}G(W)$, then $⟨v,e_i⟩=0$ for every $i$, and every generating reflection satisfiesTheir product $\sigma (c)$ therefore fixes $v$. Thus every affine Coxeter element has a nonzero fixed vector.

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