I want to prove that if the metric does not depend on some particular coordinate $x^\alpha$, The $R_{\alpha \beta}=0$ for $\alpha\neq \beta$.

My guess is we can use $\xi^\mu=\delta^\mu_\alpha$ is a killing vector and then $$ \nabla^\nu\nabla_\nu\xi_\mu=R_{\mu\nu}\xi^\nu=R_{\mu\alpha} $$

But expanding the LHS doesn't seem to give 0 for $\mu\neq \alpha$ (at least not in a trivial fashion).

PD. I have seen a couple of questions in which there are coordinate-independent solutions... I was really looking for a coordinate computation.

PD2. Sorry for the sloppy notation in the $\alpha$ index... it is a fixed index and that's why it can appear at just one side of the equation. Some people put a hat or parentheses. $\hat \alpha$, $(\alpha)$.


This is not always true. For example, let $g=y^{-2}(dx^2 + dy^2)$ be the hyperbolic metric on the upper half-plane, and make the coordinate transformation $x = u+v$, $y=v$. In $(u,v)$ coordinates, $g = v^{-2}( du^2 + 2du\,dv + 2dv^2)$, so its components don't depend on the first coordinate.

Because the Gaussian curvature is independent of coordinates, this metric has Gaussian curvature $K=-1$ just like the hyperbolic metric. And since the Ricci tensor on any $2$-manifold is just the Gaussian curvature times the metric, we have $R_{12} = K g_{12} = - v^{-2}\ne 0$.

If you assume that the metric coefficients are independent of the first coordinate, and in addition that the metric components $g_{1\beta}$ are zero when $\beta\ne 1$, then you get the result you want. In this case, the metric is (locally) a product metric on $\mathbb R \times N$ for some $(n-1)$-manifold $N$, and the result is a special case of the fact that for a product metric, all of the components of curvature vanish unless they all refer to one factor or all to the other factor. You can probably come up with a coordinate proof by showing first that the Christoffel symbols are zero if any of the indices is $1$.

EDIT: Replaced my first incorrect example with one that works (I hope!).

  • $\begingroup$ Using that metric I get the Gauss curvature is 0. And I think the result is true. I think imposing $g_{1\beta}=0 $ is far stronger than needed. $\endgroup$ – David Jaramillo Oct 22 '17 at 19:53
  • $\begingroup$ @DavidJaramillo: Oops, I guess I was too hasty. I've replaced that example by a better one above. $\endgroup$ – Jack Lee Oct 22 '17 at 21:54
  • $\begingroup$ yeah, that looks right.... hmmm I guess then the result is wrong. Will try to prove it with the extra condition you suggested ($g_{1\beta}=0$). That should be simpler.... Thanks for the help :) $\endgroup$ – David Jaramillo Oct 23 '17 at 10:10

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