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

The metric is the left-invariant metricIts right-invariant vector fields areTheir flows act by left translations, which preserve a left-invariant metric. Directly,so both are Killing vector fields and generate one-parameter isometry groups.

There is an additional Killing field. Put $y=e^a>0$ and $x=b$; thenthe 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 writtenwhich is not right invariant. Hence the answer is yes.

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