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An irreducible affine Coxeter group has a positive-semidefinite Coxeter Gram matrix with a one-dimensional radical.

A finite Coxeter group is a Coxeter group with finitely many elements. Its Coxeter Gram matrix is positive definite.

A hyperbolic Coxeter group has a nondegenerate Coxeter Gram matrix of Lorentzian signature in its standard geometric realization.

Let $\varphi =2\cos (\pi /5)=(1+\sqrt{5})/2$. In the vertex order along the displayed $3$-$5$-$3$ path, the Coxeter Gram matrix has diagonal entries $2$ and successive off-diagonal entries $-1,-\varphi ,-1$. Its leading principal determinants satisfyThe last determinant is nonzero, so the form is nondegenerate. Its negative determinant rules out positive semidefiniteness and hence also positive definiteness. Thus the answers are respectively no, no, and yes.

Solved by gpt-5.6-sol high.

For the same graph, let $e_0$ correspond to the central vertex and $e_1,\dots ,e_4$ to the leaves. The associated simply-laced Coxeter Gram matrix hasfor distinct leaves. The nonzero vectorsatisfies $(\delta ,e_i)=0$ for every basis vector. Thus the form is degenerate; in fact it is positive semidefinite with one-dimensional radical, as expected for the affine graph ${\tilde{D}}_4$.

The geometric form of a finite Coxeter group is positive definite. Since this form is degenerate, the group cannot be finite.

Solved by gpt-5.6-sol high.

Let $V$ be the real vector space with basis $(e_i{)}_{i\in I}$. The Coxeter Gram matrix defines the symmetric bilinear formIts diagonal entries are $2$. The Geometric representation of a Coxeter group is generated by the reflectionsEach has square one, and on $\mathrm{span}\{e_i,e_j\}$ the product $\sigma (x_i)\sigma (x_j)$ has order $m_{ij}$. The reflections therefore satisfy the Coxeter relations and define a group representation $\sigma :W\to GL(V)$.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.