If the metric does not depend on a coordinate $i$, then along a geodesic, $g_{i\mu}\overset{.}{x}^\mu$ is conserved

Question

In the title, $\overset{.}{x}^\mu$ is meant to mean the derivative of the geodesic with respect to the curve parameter, that is, the tangent to the geodesic. This is another of my general relativity exercises, so Einstein summation is again implied. The title basically says it all - if we have a metric that is independent of the $i$-th coordinate, then the $i$-th component of the dual of the tangent vector $\overset{.}{x}_i = g_{i\mu}\overset{.}{x}^\mu$ does not vary along the geodesic. I can show this by a brute force calculation, starting from $$ \overset{.}{x}^\alpha\nabla_\alpha(g_{i\mu}\overset{.}{x}^\mu) $$ and, after expanding the covariant derivative in terms of Christoffel symbols, seeing that all terms cancel and we are left with $0$. However, this is not satisfactory to me - I feel like the true meaning of this quantity being conserved is obfuscated by all these coordinate-based calculations. How would one show this in a coordinate-independent fashion?

Answer 1

The geodesic equation is $$ \dot x^\alpha\,\nabla_\alpha\, \dot x^\mu=0\,. $$ When $\nabla$ is metric compatible (say the Levi-Civita connection) then $$ 0=g_{i\mu}\,\dot x^\alpha\,\nabla_\alpha\, \dot x^\mu=\dot x^\alpha\,\nabla_\alpha\, (g_{i\mu}\dot x^\mu)=\dot x^\alpha\,\nabla_\alpha\, \dot x_i\,. $$ This does not require any further condition of the metric regarding its dependence on coordinates.

To see this in action recall that the flat metric in $\mathbb R^2$ in polar coordinates is $$ g_{ij}=\pmatrix{1&0\\0&r^2} $$ which depends on $x^1=r\,.$ Nonetheless, $$ \nabla_\alpha g_{ij}=\partial_\alpha g_{ij}-\Gamma^\beta_{i\alpha}g_{\beta j}-\Gamma^\beta_{\alpha j}g_{i\beta}=0 $$ for all $\alpha,i,j$ even if $\partial_\alpha g_{ij}\not=0\,.$ Namely for $\alpha=1,i=j=2$ where this equation becomes $$ 2r-\tfrac1rr^2-\tfrac1rr^2=0\,. $$