Can somebody please explain to me how the following statement is true?

The Riemann curvature tensor $R^c_{dab}$ is given by the Ricci identity $$(\nabla_a\nabla_b-\nabla_b\nabla_a)V^c\equiv R^c_{dab}V^d$$ where $\nabla_a$ denotes the covariant derivative. It is linear in $V^c$, hence may be shown by the Quotient theorem to be a tensor.

Now, I can see that the $R^c_{dab}$ is a tensor by construction -- based on the LHS of the Ricci identity. However, I don't understand how the linearity in $V^d$ comes to play.

Also, it is given that for covectors, the Ricci identity takes the form

$$(\nabla_a\nabla_b-\nabla_b\nabla_a)V_c\equiv -R^d_{cab}V_d$$

How does this follow from the Ricci identity for (contravariant) vectors?

If I write $$(\nabla_a\nabla_b-\nabla_b\nabla_a)V_c=(\nabla_a\nabla_b-\nabla_b\nabla_a)(g_{cd}V^d)$$ and in GR, the Levi-Civita connection has that the metric is covariantly constant, we have $$(\nabla_a\nabla_b-\nabla_b\nabla_a)(g_{cd}V^d)=g_{cd}(\nabla_a\nabla_b-\nabla_b\nabla_a)V^d\\=g_{cd}R^d_{eab}V^e=R_{ceab}V^e=R^d_{cab}V_d$$ Where has my minus sign gone?

I have read that you can the Ricci identity for covectors by arguing using the fact that the Levi-Civita connection is symmetric, but I don't know how they mean.

Thanks in advance for any help!

  • $\begingroup$ In this context, "symmetric" = "torsion-free" (the latter is better to use). $\endgroup$ – Yuri Vyatkin Apr 16 '13 at 4:42
  • $\begingroup$ @YuriVyatkin: Indeed. $\endgroup$ – Harrold Apr 16 '13 at 9:44

Well, the linearity is really easy: $$ (\nabla_a\nabla_b-\nabla_b\nabla_a)(f\,V^c) = f (\nabla_a\nabla_b-\nabla_b\nabla_a)V^c $$ because the connection is assumed to be torsion free: $$ (\nabla_a\nabla_b-\nabla_b\nabla_a)f=0 $$ (otherwise your "Ricci identity" won't work).

A slick way of doing this is to observe that the operator $(\nabla_a\nabla_b-\nabla_b\nabla_a)$ satisfies the product rule.

With regards to the minus sign, the problem is that the order of indices in the curvature tensor is important. It is better to write $$ (\nabla_a\nabla_b-\nabla_b\nabla_a)V^c = R_{a b}{}^c{}_d V^d $$ and then $$ \begin{align} (\nabla_a\nabla_b-\nabla_b\nabla_a)V_c & = (\nabla_a\nabla_b-\nabla_b\nabla_a)(g_{cd}V^d) \\ &= g_{cd}(\nabla_a\nabla_b-\nabla_b\nabla_a)V^d\\ & =g_{cd} R_{a b}{}^d{}_e V^e \\ &= R_{a b d e} V^e = - R_{a b e d} V^e = - R_{a b}{}^c{}_d V_c \end{align} $$

(I have learned all this from R. Wald's "General relativity")

  • $\begingroup$ Thank you very very much, Yuri! (Extra thanks for the reference -- I have been looking for a good reference book.) :) $\endgroup$ – Harrold Apr 16 '13 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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