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The Lie derivative of a connection is the tensorAn isometry preserves the Levi-Civita connection, so ${\mathcal{L}}_K\nabla =0$ for a Killing vector field.

A Killing vector field $K$ obeys ${\mathcal{L}}_{K}g=0$, or equivalently ${\nabla }_{(a}{K}_{b)}=0$. Along an affinely parametrized geodesic with tangent $U$,The second term vanishes by the geodesic equation, while the first contracts the symmetric tensor $U^a{U}^b$ with the antisymmetric part selected by the Killing equation. Thus $g(K,U)$ is a geodesic conserved quantity from a Killing vector. In Minkowski spacetime, translational Killing fields give conserved energy-momentum and rotational or boost Killing fields give the corresponding angular-momentum and boost charges; the same construction works for any spacetime isometry.

Solved by gpt-5.6-sol high.

The flow of a Killing vector field consists of local isometries. Isometries preserve both the metric and its unique torsion-free metric-compatible Levi-Civita connection, hence ${\mathcal{L}}_K\nabla =0$. Set $X=K$ and $Y=Z=U$ in part (b). Since ${\nabla }_{U}U=0$,and antisymmetry of the first two curvature arguments givesThis is the geodesic deviation equation with connecting field $K$: applying the isometry to the original geodesic produces a neighboring geodesic, and curvature determines their relative acceleration.

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The four independent Killing vector fields are the stationary field ${\partial }_t$ and the three generators of spatial rotations on the two-spheres. In the Schwarzschild interior, $f(r)<0$, soEvery rotational Killing field is tangent to the positive-definite round-sphere metric. All four fields are tangent to a surface of constant $r$, whose induced metricis positive definite. Consequently every nonzero linear combination of the Killing fields is spacelike wherever it does not vanish. A pure rotational Killing field can vanish on its rotation axis, but it is nowhere timelike.

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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.

Solved by gpt-5.6-sol high.