Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 3

Table of contents
- a 3
  - Solution a
- b 3
  - Solution b
- c 3
  - Solution c
- d 3
  - Solution d

a

Solution

For a matrix Lie group, the left-invariant Maurer-Cartan form is

ρ = g^{- 1} d g .

Differentiating

g^{- 1} g = I

gives

d (g^{- 1}) = - g^{- 1} (d g) g^{- 1}

. Hence

d ρ = d (g^{- 1}) \land d g = - g^{- 1} d g \land g^{- 1} d g = - ρ \land ρ,

and therefore

d ρ + ρ \land ρ = 0 .

Solved by gpt-5.6-sol high.

b

Solution

Substitute

ρ = σ^{α} T_{α}

into the Maurer-Cartan equation. Antisymmetry of the wedge product gives

ρ \land ρ = \frac{1}{2} σ^{α} \land σ^{β} [T_{α}, T_{β}] = \frac{1}{2} c^{γ}_{α β} σ^{α} \land σ^{β} T_{γ} .

Equating coefficients of

T_{γ}

yields

d σ^{γ} = - \frac{1}{2} α, β \sum c^{γ}_{α β} σ^{α} \land σ^{β},

so

f^{γ}_{α β} = - c^{γ}_{α β} /2

.

Solved by gpt-5.6-sol high.

c

Solution

Represent the real affine group by

g (a, b) = (e^{a} 0 b 1) .

Matrix multiplication reproduces

(a, b) (a^{'}, b^{'}) = (a + a^{'}, b + e^{a} b^{'}) .

The Maurer-Cartan form is

g^{- 1} d g = (d a 0 e^{- a} d b 0),

so a basis of left-invariant one-forms is

σ^{1} = d a, σ^{2} = e^{- a} d b .

The dual left-invariant vector fields are

E_{1} = \partial_{a}, E_{2} = e^{a} \partial_{b} .

Indeed

σ^{i} (E_{j}) = δ_{j}^{i}

, and left translation preserves the one-forms and vector fields.

Solved by gpt-5.6-sol high.

d

Solution

The metric is the left-invariant metric

g = d a^{2} + e^{- 2 a} d b^{2} .

Its right-invariant vector fields are

R_{1} = \partial_{a} + b \partial_{b}, R_{2} = \partial_{b} .

Their flows act by left translations, which preserve a left-invariant metric. Directly,

L_{R_{1}} g = 0, L_{R_{2}} g = 0,

so both are Killing vector fields and generate one-parameter isometry groups.

There is an additional Killing field. Put

y = e^{a} > 0

and

x = b

; then

g = \frac{d x ^{2} + d y ^{2}}{y ^{2}},

the hyperbolic plane of constant curvature

- 1

. Its isometry algebra is three-dimensional, whereas the space of right-invariant fields here is two-dimensional. For example, the third independent Killing field can be written

(b^{2} - e^{2 a}) \partial_{b} + 2 b \partial_{a},

which is not right invariant. Hence the answer is yes.

Solved by gpt-5.6-sol high.

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