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I am relatively used to the coordinate free expression of the Riemann tensor: $$ R(X, Y)Z=\nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X, Y]} Z, $$ where $\nabla$ is the Levi-Civita connection on a (pseudo)-Riemannian manifold and $X, Y, Z$ are contravariant vector fields. This $R$ is a tensor field with $3$ covariant indices and $1$ contravariant index.

Problem is, I am having a hard time figuring out why the following formula (taken from Wald, "General Relativity", §3.2) defines the same tensor field: $$ R_{abc}{}^d\omega_d=\nabla_a\nabla_b\omega_c - \nabla_b\nabla_a \omega_c,$$ where $\omega$ is an arbitrary covariant vector field. (This formula uses the abstract index notation, but I think it can be safely read by interpreting indices the usual way).

Can you show me how to prove equivalence or point me to some reference?

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1 Answer 1

up vote 7 down vote accepted

It is a general theorem that to establish a tensor identity involving Lie brackets, one actually only needs to consider the case where all Lie brackets vanish. This is because we can always find a basis of commuting vector fields by taking any coordinates and using the coordinate vector fields $\partial/\partial x^\mu$, and if a tensor identity holds for these, it holds for all vector fields by linearity. Then equality between the two definitions is fairly clear: ignoring the last term in the coordinate-free expression, the second definition is just the first in index notation.

More concretely in this particular case, we can take $X = X^\mu \partial_\mu, Y = Y^\mu \partial_\mu$. Note that $X^\mu, Y^\mu$ are scalars. Let us abbreviate $\nabla_{\partial_\mu}$ by $\nabla_\mu$. Then $$\nabla_X \nabla_Y Z = \nabla_X (Y^\nu \nabla_\nu Z) = X^\mu (\nabla_\mu Y^\nu)(\nabla_\nu Z) + X^\mu Y^\nu \nabla_\mu\nabla_\nu Z. $$ Here I have used $C^\infty$-linearity of the covariant derivative in the subscript and the Leibniz rule. Of course for the second term we just swap $X \leftrightarrow Y$, $$\nabla_Y \nabla_X Z = \nabla_Y (X^\nu \nabla_\nu Z) = Y^\mu (\nabla_\mu X^\nu)(\nabla_\nu Z) + Y^\mu X^\nu \nabla_\mu\nabla_\nu Z. $$ We see that $$\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z = Y^\mu X^\nu(\nabla_\mu\nabla_\nu - \nabla_\nu\nabla_\mu)Z + (X^\mu \nabla_\mu Y^\nu - Y^\mu \nabla_\mu X^\nu)(\nabla_\nu Z).$$ But the second term is precisely $\nabla_{[X,Y]} Z$, so $$R(X,Y)Z = X^\mu Y^\nu R_{\mu\nu}{}^c{}_d Z^d.$$ This differs from Wald's definition, but it's just a matter of moving indices, $$R_{\mu\nu}{}^c{}_d Z^d = R_{\mu\nu}{}^{cd} Z_d$$ and contracting with the metric, $$R_{\mu\nu c}{}^d Z_d = g_{c\alpha}(\nabla_\mu\nabla_\nu - \nabla_\nu\nabla_\mu) Z^\alpha = (\nabla_\mu\nabla_\nu - \nabla_\nu\nabla_\mu) Z_c$$ where the last step is that metric contractions commute with $\nabla$, since we are using the Levi-Civita connection.

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