The point is that, with your second equation, you are dealing with in a local coordinate patch, say an open set U⊂M equipped with coordinates xμ≡x1,x2,x3,x4. Therefore, if p∈U, you can handle two bases of the tangent space TpM. One is made of (pseudo)orthonormal vectors ea, a=1,2,3,4 and the other is the one associated with the coordinates ∂∂xμ|p, μ=1,2,3,4. The metric at p reads:
gp=ηabωa⊗ωb
where, by definition, the co-vectors
ωa∈T∗pM (defining another pseudo-orthonormal tetrad but in the cotangent space at
p) satisfy
ωa(eb)=δab.(0)
Correspondingly you have:
ωa=ωaμdxμ|p,(1)
and
Ω:=[ωaμ] is a
4×4 invertible matrix. Invertible because it is the transformation matrix between two bases of the same vector space (
T∗pM).
Similarly
ea=eμa∂∂xμ|p,(2)
where
E:=[eaμ] is a
4×4 invertible matrix. It is an elementary exercise to prove that, in view of (0):
E=Ω−1t,(3)
so you can
equivalently write
eμa=(ω−1)μa
where the
transposition operation in (3) is now apparent from the fact that we have swapped the positions of Greek and Latin indices (compare with (1)).
An identity as yours:
Rμνλτ=(ω−1)μaωbνωcλωdτRabcd
is understood as a trivial change of basis relying upon (1) and (2). It could equivalently be written down as:
Rμνλτ=eμaωbνωcλωdτRabcd.
This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Valter Moretti