The <Killing equation> gives $\nabla_\mu K^\mu=0$ and $\nabla_{[\mu}K_{\lambda]}=\nabla_\mu K_\lambda$. Commuting derivatives once and contracting the <Ricci identity> gives
$$
\nabla^\lambda\nabla_\mu K_\lambda=R_{\mu\beta}K^\beta.
$$
Therefore, using the <contracted Bianchi identity> and symmetry of $R_{\mu\beta}$,
$$
\begin{aligned}
2\nabla^\mu\nabla^\lambda\nabla_{[\mu}K_{\lambda]}
&=2\nabla^\mu(R_{\mu\beta}K^\beta)\\
&=K^\beta\nabla_\beta R
+2R_{\mu\beta}\nabla^\mu K^\beta\\
&=K^\beta\nabla_\beta R.
\end{aligned}
$$
Since $\nabla_{[\mu}K_{\lambda]}$ is antisymmetric, swapping the two contracted index names shows that the left side equals its derivative-antisymmetrized form:
$$
\boxed{K^\alpha\nabla_\alpha R
=2\nabla^{[\mu}\nabla^{\lambda]}\nabla_{[\mu}K_{\lambda]}.}
$$
Now set $A_{\mu\lambda}=\nabla_{[\mu}K_{\lambda]}$. The right side is the contracted commutator $[\nabla^\mu,\nabla^\lambda]A_{\mu\lambda}$. By the <curvature commutator on a covariant tensor>, it is a sum of contractions of the symmetric <Ricci tensor> with the antisymmetric tensor $A$, and hence vanishes. Thus
$$
\boxed{K^\alpha\nabla_\alpha R=0,}
$$
so the <Killing vector preserves scalar curvature>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 37D / d / Solution

Every <Killing vector field> $K$ preserves the <scalar curvature>: $\mathcal L_KR=K^\alpha\nabla_\alpha R=0$. One index proof rewrites this derivative as a doubly contracted curvature commutator acting on the antisymmetric tensor $\nabla_{[\mu}K_{\lambda]}$; both terms vanish by symmetry of the <Ricci tensor>.

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