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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!

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  • $\begingroup$ 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
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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")

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  • $\begingroup$ 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

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