Non-diagonal elements of the Schwarzchild metric

Question

The Schwarzchild metric is the most general spherically symmetric, vacuum solution of the Einstein field equations.

I was wondering if there was a simple argument to explain why the Schwarzchild metric is diagonal in the spherical coordinate system, i.e. of the form

$$ds^2 = dt^2 + \cdots d\theta^2 + \cdots d\phi^2 + \cdots dr^2.$$

This Wikipedia article gives a really simple explanation which seems false.

(When you write the transformation law for $ g_{\mu 4} $, it should be understood: $$ g'_{\mu 4} (x') = \frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^4}g_{\alpha\beta}(x)= -g_{\mu 4}(x) ,$$ while at the same time the invariance tells you that $$ g'_{\mu 4}(x') = g_{\mu 4}(x') .$$ This leads to the conclusion that $$ g_{\mu 4}(x') = -g_{\mu 4}(x) ,$$ but I fail to see how to go further in the reasoning without any aditionnal assumption.)

Other derivations either start from the diagonal form or are much more complicated.

It's probably a dumb question, but I fail to see a simple argument.

Answer 1

OP is right: For a generic pseudo-Riemannian manifold $(M;g)$, there does not necessarily exist an open coordinate neighborhoods $U\subseteq M$, where the metric $g_{|U}$ is on diagonal form. Fermi normal coordinates always ensure a diagonal form along a geodesic $\gamma$ (but not necessarily in the ambient spacetime outside the geodesic). However, the Schwarzschild geometry has Killing symmetries that ensure that diagonal metrics exist in open neighborhoods.