Because the metric coefficients are independent of $t$ and $y$,
$$
\partial_t,qquad \partial_y
$$
are <Killing vector fields>. The maps
$$
\phi_s(t,x,y)=(t,x+s,e^{-s}y)
$$
satisfy $\phi_s\circ\phi_r=\phi_{s+r}$. Moreover, $dx$ is unchanged and both $e^xdy$ and $dt+\sqrt2e^xdy$ are invariant, so $\phi_s^*g=g$. Differentiating at $s=0$ produces the third Killing field
$$
\boxed{K=\partial_x-y\partial_y}.
$$
Given two points, first use $\phi_s$ 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>.

Solution

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

{wiki}

A homogeneous space is a space on which a group acts transitively. After choosing a point, a transitive Lie-group action identifies the space with a quotient $G/H$ by the point stabilizer.


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