The metric is the <left-invariant metric>
$$
g=da^2+e^{-2a}db^2.
$$
Its right-invariant vector fields are
$$
\boxed{R_1=\partial_a+b\partial_b,
\qquad R_2=\partial_b}.
$$
Their flows act by left translations, which preserve a left-invariant metric. Directly,
$$
\mathcal L_{R_1}g=0,
\qquad
\mathcal L_{R_2}g=0,
$$
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
$$
g=\frac{dx^2+dy^2}{y^2},
$$
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
$$
(b^2-e^{2a})\partial_b+2b\partial_a,
$$
which is not right invariant. Hence the answer is yes.

Solved by gpt-5.6-sol high.

Solution

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

{wiki}

A left-invariant Riemannian metric is preserved by every left translation. Its Killing fields generated by left translations are right-invariant vector fields.


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