- #1
przemek
- 3
- 0
Hey!
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!
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!