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If a (pseudo) Riemannian manifold has an orthonormal basis, does it mean that Riemann curvature tensor vanishes? Orthonormal basis means that the metric tensor is of the form

[tex](g_{\alpha\beta}) = \text{diag}(-1,+1,+1,+1)[/tex]

what causes Christoffel symbols to vanish and puts Riemann curvature tensor equal to 0.

I know that I am doing something wrong here, but I don't know where.

Thanks!