I am studying general relativity out of Sean Carroll's textbook, and I had trouble proving the following assertion (which is more about differential geometry than physics per se). Suppose we have a metric $g_{\mu \nu}$ in some coordinate representation on a manifold such that $g$ is independent of a special coordinate $\sigma_*$.
$$\partial_{\sigma_*} g_{\mu \nu} = 0$$
I would like to show that the vector field along the coordinate $\partial_{\sigma_*}$ is in fact a Killing vector (field). In particular, I would like to demonstrate this by showing that
$$\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0$$
where $K_\mu = (\partial_{\sigma_*})_\mu = \delta_{\sigma_* \mu}$. Note that $\nabla$ is given by the Levi-Cevita connection.
Since the components $K_\mu$ are clearly independent of our coordinates, partial derivatives of $K_\mu$ must vanish, leaving
$$\nabla_\mu K_\nu = \Gamma^{\lambda}_{\mu\nu} K_\lambda = \Gamma^{\sigma_*}_{\mu \nu}.$$
Performing the same calculation for $\nabla_\nu K_\mu$ and invoking metric compatibility to assert $\nabla$ is torsion free ($\Gamma^{\sigma_*}_{\mu\nu} = \Gamma^{\sigma_*}_{\nu\mu})$,
$$\nabla_\mu K_\nu + \nabla_\nu K_\mu = 2 \Gamma_{\mu\nu}^{\sigma_*}.$$
It remains to show that $\Gamma^{\sigma_*}_{\mu\nu} = 0$, and this is not clear to me without stronger assumptions. Writing $\Gamma$ in terms of the metric,
$$\Gamma^{\sigma_*}_{\mu\nu} = \frac{1}{2} g^{\sigma_* \lambda}\left(\partial_\mu g_{\lambda\nu} + \partial_\nu g_{\lambda\mu}- \partial_\lambda g_{\mu\nu}\right).$$
It is not clear to me that this is necessarily zero, and I feel required to impose more stringent conditions on the metric. For example, I could assert that $g_{\lambda \sigma_*} = 0$ unless $\lambda = \sigma_*$. In this case, I would find most of the Christoffel symbols are zero except perhaps
$$2 \Gamma^{\sigma_*}_{\sigma_* \nu} = g^{\sigma_* \sigma_*} \partial_\nu g_{\sigma_* \sigma_*}$$
for $\nu \neq \sigma_*$. Even in this special case, the answer may be nonzero.
What am I missing? Is the thing I am trying to prove actually false, and if so is there a correct version of the idea I am trying to get at (coordinate-independent metric implies "symmetry" through Killing vector)?
Best Answer
I'll omit the star for typing convenience. The vector field is $K=\frac{\partial}{\partial x^{\sigma}}$. So, $K^\mu=\delta^{\mu}_{\sigma}$, meaning the covector components are $K_{\mu}=g_{\mu\nu}K^{\nu}=g_{\mu\nu}\delta^{\nu}_{\,\sigma}=g_{\mu\sigma}$. This is your mistake.
Now, if we compute, we see that \begin{align} \nabla_{\mu}K_{\nu}&=\frac{\partial K_{\nu}}{\partial x^{\mu}}-\Gamma_{\mu\nu}^{\alpha}K_{\alpha}\\ &=\frac{\partial g_{\nu\sigma}}{\partial x^{\mu}}-g_{\alpha\sigma}\Gamma^{\alpha}_{\mu\nu}\\ &=\frac{\partial g_{\nu\sigma}}{\partial x^{\mu}}-g_{\alpha\sigma}\cdot \frac{1}{2}g^{\alpha\beta}\left(\frac{\partial g_{\beta\mu}}{\partial x^{\nu}}+ \frac{\partial g_{\nu\beta}}{\partial x^{\mu}} - \frac{\partial g_{\mu\nu}}{\partial x^{\beta}}\right)\\ &=\frac{\partial g_{\nu\sigma}}{\partial x^{\mu}}-\frac{1}{2}\left(\frac{\partial g_{\sigma\mu}}{\partial x^{\nu}}+ \frac{\partial g_{\nu\sigma}}{\partial x^{\mu}} - \underbrace{\frac{\partial g_{\mu\nu}}{\partial x^{\sigma}}}_{=0}\right)\\ &=\frac{1}{2}\left(\frac{\partial g_{\nu\sigma}}{\partial x^{\mu}}-\frac{\partial g_{\sigma\mu}}{\partial x^{\nu}}\right) \end{align} Because of the symmetry of $g$, this expression is skew-symmetric in $\mu$ and $\nu$, so $\nabla_{\mu}K_{\nu}+\nabla_{\nu}K_{\mu}=0$.
Alternatively, you just compute the Lie derivative directly and see that it vanishes. Write $g=g_{\mu\nu}dx^{\mu}\otimes dx^{\nu}$, and use the fact that the Lie derivative satisfies the product rule: \begin{align} L_Kg&=(L_K(g_{\mu\nu}))\,dx^{\mu}\otimes dx^{\nu}+g_{\mu\nu} (L_K(dx^{\mu}))\otimes dx^{\nu}+g_{\mu\nu}\,dx^{\mu}\otimes (L_K(dx^{\nu})).\tag{1} \end{align} Now, for the last two terms, we note that Lie derivatives commute with exterior derivatives and the fact that $K=\frac{\partial}{\partial x^{\sigma}}$ to get \begin{align} L_K(dx^{\mu})=d(L_K(x^{\mu}))=d\left(\frac{\partial x^{\mu}}{\partial x^{\sigma}}\right)=d(\delta^{\mu}_{\sigma})=0, \end{align} since the $d$ of a constant function is $0$. So, the last two terms of $(1)$ vanish. For the first term, $L_K(g_{\mu\nu})=\frac{\partial g_{\mu\nu}}{\partial x^{\sigma}}=0$ by hypothesis. Hence, we obtain $L_Kg=0$, proving that $K=\frac{\partial}{\partial x^{\sigma}}$ is indeed a Killing vector field.