Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 111 / 2

Table of contents
- a 2
  - Solution a
- b 2
  - Solution b
- c 2
  - Solution c
- d 2
  - Solution d

a

Solution

Let

V

be the real vector space with basis

(e_{i})_{i \in I}

. The Coxeter Gram matrix defines the symmetric bilinear form

⟨ e_{i}, e_{j} ⟩ = G (W)_{ij} = - 2 cos (\frac{π}{m _{ij}}) .

Its diagonal entries are

2

. The Geometric representation of a Coxeter group is generated by the reflections

σ (x_{i}) (v) = v - ⟨ v, e_{i} ⟩ e_{i} .

Each has square one, and on

span {e_{i}, e_{j}}

the product

σ (x_{i}) σ (x_{j})

has order

m_{ij}

. The reflections therefore satisfy the Coxeter relations and define a group representation

σ : W \to G L (V)

.

Solved by gpt-5.6-sol high.

b

Solution

The Coxeter graph has vertex set

I

. Distinct vertices

i, j

are joined precisely when

m_{ij} \geq 3

, and the edge is labelled

m_{ij}

when

m_{ij} > 3

; the customary unlabelled edge therefore means

m_{ij} = 3

. The Coxeter system is an Irreducible Coxeter system precisely when this graph is connected.

Solved by gpt-5.6-sol high.

c

Solution

Put

a = 2 cos (\frac{π}{m _{ij}}) .

After ordering

I_{1}

before

I_{2}

, the only nonzero off-diagonal entries between the two blocks occur at

(i, j)

and

(j, i)

, where they equal

- a

. Thus

G (W) = (G (W_{1}) - a, e_{j} e_{i}^{T} - a, e_{i} e_{j}^{T} G (W_{2})),

where the coordinate vectors in the two blocks are understood. Expanding the determinant according to whether neither or both cross-block entries are selected gives

det G (W) = det G (W_{1}) det G (W_{2}) - a^{2} det G (W_{1}^{'}) det G (W_{2}^{'}) .

The minus sign is the sign of the transposition pairing the two cross-block entries. This formula remains valid when either diagonal block is singular, so no inverse or Schur complement is needed.

Solved by gpt-5.6-sol high.

d

Solution

All edges in the Type E(p,q,k) Coxeter graph are unlabelled, so the factor

a^{2}

in part c is one. Separate the arm of length

p

from the central vertex. The remaining two arms form a type

A_{q + k + 1}

chain, while deleting the central vertex leaves the disjoint type

A_{q}

and type

A_{k}

chains. Using

det G (A_{r}) = r + 1

in the formula from part c yields

det G (E (p, q, k)) = (p + 1) (q + k + 2) - p (q + 1) (k + 1) = (p + 1) (q + 1) (k + 1) (\frac{1}{p + 1} + \frac{1}{q + 1} + \frac{1}{k + 1} - 1) .

The associated bilinear form is degenerate exactly when

\frac{1}{p + 1} + \frac{1}{q + 1} + \frac{1}{k + 1} = 1.

The positive-integer solutions of

1/ a + 1/ b + 1/ c = 1

, up to permutation, are

(3, 3, 3)

,

(2, 4, 4)

, and

(2, 3, 6)

. Consequently the degenerate arm-length triples are the permutations of

(2, 2, 2), (1, 3, 3), (1, 2, 5) .

For every other allowed

(p, q, k)

the determinant is nonzero, so the form is nondegenerate.

Solved by gpt-5.6-sol high.

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