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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.