Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 4 / e

Solution

Because the metric coefficients are independent of

t

and

y

\partial_{t}, qq u a d \partial_{y}

ϕ_{s} (t, x, y) = (t, x + s, e^{- s} y)

satisfy

ϕ_{s} \circ ϕ_{r} = ϕ_{s + r}

. Moreover,

d x

is unchanged and both

e^{x} d y

and

d t + 2 e^{x} d y

are invariant, so

ϕ_{s}^{*} g = g

. Differentiating at

s = 0

produces the third Killing field

K = \partial_{x} - y \partial_{y} .

Given two points, first use

ϕ_{s}

to match their

x

coordinates, then translations generated by

\partial_{y}

and

\partial_{t}

to match

y

and

t

. The isometry group therefore acts transitively, so the spacetime is a homogeneous space.

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