Killing temperature normalization and rigidity

Differentiating $g(\xi ,\xi )$ with the metric-compatible connection gives $\xi (N_{\xi })=0$, since ${\xi }^a{\xi }^b{\mathrm{\nabla }}_a{\xi }_b=0$ by the Killing equation. The same equation gives ${\xi }^a{\mathrm{\nabla }}_a{\xi }_b=-\frac{1}{2}{\mathrm{\nabla }}_{b}g(\xi ,\xi )=N_{\xi }{\mathrm{\nabla }}_b{N}_{\xi }$. Therefore $a_b=u^a{\mathrm{\nabla }}_a{u}_b={\mathrm{\nabla }}_b\log N_{\xi }$. Also $\theta ={\mathrm{\nabla }}_a({\xi }^a\mathrm{/}{N}_{\xi })=0$ because ${\mathrm{\nabla }}_a{\xi }^a=0$ and $\xi (N_{\xi })=0$. The symmetric derivative is ${\mathrm{\nabla }}_{(a}{u}_{b)}=-u_{(a}{\mathrm{\nabla }}_{b)}\log N_{\xi }$; projecting both indices with $h$ gives zero, so ${\sigma }_{ab}=0$.

Put $f:=(k_B{N}_{\xi }T{)}^{-1}>0$, so $\beta =f\xi$. If $\beta$ is Killing, both $f\xi$ and $\xi$ have zero symmetric covariant derivative, and hence ${\xi }_{(b}{\mathrm{\nabla }}_{a)}f=0$. Contracting with $u^a{u}^b$ gives $u(f)=0$ because $\xi =N_{\xi }u$ and $N_{\xi }>0$; contracting with $u^a{h}^b{}_c$ gives $h^b{}_c{\mathrm{\nabla }}_{b}f=0$. The decomposition of a covector into its $u$ and rest-space parts now gives $df=0$.

In a convex coordinate ball in $O$, join any two coordinate points by a straight segment. The derivative of $f$ along that segment is zero by the covector decomposition just proved, so [F2] makes $f$ constant on the ball. Thus $f$ is locally constant on $O$. Each level set of $f$ and its complement are open; connectedness of $O$ therefore makes $f$ one positive constant. Equivalently $N_{\xi }T=T_K$ for a constant $T_K>0$, and then $\beta =\xi \mathrm{/}(k_B{T}_K)$.

Conversely, if $N_{\xi }T=T_K$ is constant, then $\beta =\xi \mathrm{/}(k_B{T}_K)$ is a constant multiple of the Killing field $\xi$ and is Killing. Under $\xi \mapsto b\xi$, $N_{\xi }\mapsto b{N}_{\xi }$ and $T_K\mapsto b{T}_K$, whereas $u=\xi \mathrm{/}{N}_{\xi }$ and $\beta =\xi \mathrm{/}(k_B{T}_K)$ remain fixed. These identities prove the converse and the normalization statement.