2. (a) By considering (-)(U^c), use the Ricci identity for vectors and the fact that covariant d

Step 1: Start with the expression (VaVa - VoVa)Tcd.

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2. (a) By considering ($\nabla_a \nabla_b - \nabla_b \nabla_a$)(U$^c \Omega_c$), use the Ricci identity for vectors and the fact that covariant derivatives commute on scalars to obtain the Ricci identity for 1-forms ($\nabla_a \nabla_b - \nabla_b \nabla_a$) $\Omega_c = -R^e_{cab} \Omega_e$. (b) Show similarly that for type ($^2_0$) tensors ($\nabla_a \nabla_b - \nabla_b \nabla_a$)T$^{cd} = R^c_{eab} T^{ed} + R^d_{eab} T^{ce}$ and consequently deduce that if F$^{ab} = -F^{ba}$ is an antisymmetric tensor $\nabla_a \nabla_b F^{ab} = 0$, so that $\nabla_a \nabla_b T^{ab} = \nabla_a \nabla_b T^{ba}$ for any tensor T$^{ab}$.

          2. (a) By considering ($\nabla_a \nabla_b - \nabla_b \nabla_a$)(U$^c \Omega_c$), use the Ricci identity for vectors and the fact that covariant derivatives commute on scalars to obtain the Ricci identity for 1-forms

($\nabla_a \nabla_b - \nabla_b \nabla_a$) $\Omega_c = -R^e_{cab} \Omega_e$.

(b) Show similarly that for type ($^2_0$) tensors

($\nabla_a \nabla_b - \nabla_b \nabla_a$)T$^{cd} = R^c_{eab} T^{ed} + R^d_{eab} T^{ce}$

and consequently deduce that if F$^{ab} = -F^{ba}$ is an antisymmetric tensor $\nabla_a \nabla_b F^{ab} = 0$, so that $\nabla_a \nabla_b T^{ab} = \nabla_a \nabla_b T^{ba}$ for any tensor T$^{ab}$.

2. (a) By considering (-)(U^c), use the Ricci identity for vectors and the fact that covariant derivatives commute on scalars to obtain the Ricci identity for 1-forms

(-) = -R^ecab.

(b) Show similarly that for type (^20) tensors

(-)T^cd = R^ceab T^ed + R^deab T^ce

and consequently deduce that if F^ab = -F^ba is an antisymmetric tensor F^ab = 0, so that T^ab = T^ba for any tensor T^ab.

Added by Jeffrey M.

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