We show here that coordinate directions in which the metric (line element) doesn't change always correspond to Killing vectors.

Theorem: Suppose that $\{x=y^0,y^1,…\}$ are orthogonal coordinates, so that 1) \begin{equation} d\rr = h\,dx\,\xhat + \sum\limits_i h_i\,dy^i\,\yhat{}^i \end{equation} and suppose further that the coefficients $h=h_0,h_1,…$ do not depend on $x$, that is, suppose that \begin{equation} \Partial{h}{x} = 0 = \Partial{h_i}{x} \end{equation} Then $\XX = h\xhat$ is a Killing vector.

Proof: 2)

Our orthonormal basis of 1-forms is $\{\sigma^x=h\,dx,\sigma^i=h_i\,dy^i\}$. The structure equation for $d\sigma^x$ is \begin{align} 0 &= d\sigma^x + \sum\limits_i\omega^x{}_i\wedge\sigma^i \nonumber\\ &= dh \wedge dx + \sum\limits_i\omega^x{}_i\wedge h_i\,dy^i \label{gstruct} \end{align} Writing \begin{equation} \omega^x{}_i = \Gamma^x{}_{ix}\,\sigma^x + \sum\limits_j\Gamma^x{}_{ij}\,\sigma^j \end{equation} and collecting terms not involving $dx$, we obtain \begin{equation} \sum\limits_{i,j} \Gamma^x{}_{ij}\,\sigma^j\wedge\sigma^i = 0 \label{gsym} \end{equation}

On the other hand, using the results (and notation) of Uniqueness, we have \begin{align} 2 \Gamma_{xij} &= g(d\sigma_x,\sigma_i\wedge\sigma_j) - g(d\sigma_j,\sigma_x\wedge\sigma_i) + g(d\sigma_i,\sigma_j\wedge\sigma_x) \nonumber\\ &= g(d\sigma_x,\sigma_i\wedge\sigma_j) \label{gform} \end{align} where \begin{equation} \Gamma_{xij} = \epsilon\,\Gamma^x{}_{ij} \end{equation} and \begin{equation} \epsilon = \xhat\cdot\xhat = \pm1 \end{equation} (The last two terms in the first equality in (\ref{gform}) vanish, because $d\sigma_i=dh_i\wedge dy^i$ contains no $dx$ terms by assumption, and is hence orthogonal to any 2-form containing $\sigma^x$.) Since the final expression in (\ref{gform}) is antisymmetric in the indices $i$ and $j$, so is $\Gamma^x{}_{ij}$, that is, \begin{equation} \Gamma^x{}_{ji} = -\Gamma^x{}_{ij} \end{equation} But (\ref{gsym}) implies that $\Gamma^x{}_{ij}$ is symmetric in $i$ and $j$, and must therefore vanish.

Thus, $\omega^x{}_i$ is proportional to $\sigma^x=h\,dx$, and we can therefore conclude that \begin{equation} \omega^x{}_i = \frac{1}{h_i}\Partial{h}{y^i}\,dx \end{equation}

Since \begin{equation} d\xhat\cdot\xhat = \frac12 d(\xhat\cdot\xhat) = 0 \end{equation} we have \begin{align} d\xhat\cdot d\rr &= 0 + \sum\limits_i h_i\,dy^i\,d\xhat\cdot\yhat{}^i = \sum\limits_i h_i\,dy^i\,\omega_{ix} \nonumber\\ &= -\epsilon\sum\limits_i h_i\,dy^i\,\omega^x{}_i = -\epsilon\,dh\,dx \end{align} and of course \begin{equation} \xhat\cdot d\rr = \epsilon\,h\,dx \end{equation}

Putting this all together, we have \begin{align} d\XX\cdot d\rr &= d(h\,\xhat)\cdot d\rr \nonumber\\ &= dh\,\xhat\cdot d\rr + h\,d\xhat\cdot d\rr \\ &= \epsilon\,h\,dh\,dx - \epsilon\,h\,dh\,dx \nonumber\\ &= 0 \nonumber \end{align}

Thus, $\XX$ is indeed a Killing vector, as claimed.