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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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You got it right up to $$ ...=R_{ceab}V^e=-R_{ecab}V^e=-R^d_{\cdot cab}V_d. $$ There is a difference between the first index raised ($R^c_{\cdot dab}$) and the second idex raised ($R^{\,d}_{c\cdot ab}$).

As I understand it, the statement is that from the LHS of the Ricci identity you only know that $T^c_{ab}(V)=R^c_{dab}V^d$ is a tensor for any $V$. And then you use the Quotient theorem to deduce that $R^c_{dab}$ is a tensor.

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Thank you, Peter. – Harrold Apr 16 '13 at 0:15

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