Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 111 / 2

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.

The displayed simply-laced tree has arms of lengths $2$, $1$, and $3$ from its trivalent vertex, so it is the finite type $E_7$ Coxeter graph. Its Gram matrix is the $E_7$ Cartan matrix, which has positive leading principal minors in a leaf-removal ordering and determinant $2$. By Sylvester's criterion it is positive definite. It is therefore positive semidefinite and nondegenerate as well: the three answers are yes, yes, and yes.

Solved by gpt-5.6-sol high.

For the four-cycle, the quadratic form isIt is nonnegative, but it vanishes on the nonzero vector $(1,1,1,1)$. Equivalently, the Gram eigenvalues are $0,2,2,4$. The form is positive semidefinite, not positive definite, and degenerate: the three answers are no, yes, and no. This is the affine Coxeter graph ${\tilde{A}}_3$.

Solved by gpt-5.6-sol high.

The exchange condition for a Coxeter group says that if $w=s_1\cdots s_n$ is reduced and $s$ is simple with $\ell (ws)<\ell (w)$, thenfor some $j$. The Matsumoto theorem says that any two reduced expressions for the same element are connected by braid moves. Together they imply the Tits word reduction theorem: a nonreduced word can be transformed by braid moves until two equal adjacent generators can be cancelled.

For the displayed four-armed graph, call the central generator $s$ and the leaves $a,b,c,d$. Different leaves commute, while each leaf $t$ satisfies $sts=tst$. Consider the wordBetween successive occurrences of $s$, the intervening leaf sets alternate between $\{a,b\}$ and $\{c,d\}$. Commuting the two leaves in one block never puts the same leaf on both sides of an $s$, so no length-three braid $tst\leftrightarrow sts$ is ever available. The only possible braid moves are those leaf commutations, and they cannot create adjacent equal letters. Tits reduction therefore shows that $u_N$ is reduced. Since $\ell (u_N)=6N$ is unbounded, the group is infinite.

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.