1

This identity can be generalized to get the commutators for two covariant derivatives of arbitrary tensors as follows $ \begin{align} T^{\alpha_1 \cdots \alpha_r}{}_{\beta_1 \cdots \beta_s ; \gamma \delta} - T^{\alpha_1 \cdots \alpha_r}{}_{\beta_1 \cdots \beta_s ; \delta \gamma} = \, & - R^{\alpha_1}{}_{\rho \gamma \delta} T^{\rho \alpha_2 \cdots \alpha_r}{}_{\beta_1 \cdots \beta_s} - \cdots - R^{\alpha_r}{}_{\rho \gamma \delta} T^{\alpha_1 \cdots \alpha_{r-1} \rho}{}_{\beta_1 \cdots \beta_s} \\ & + \, R^\sigma{}_{\beta_1 \gamma \delta} T^{\alpha_1 \cdots \alpha_r}{}_{\sigma \beta_2 \cdots \beta_s} + \cdots + R^\sigma{}_{\beta_s \gamma \delta} T^{\alpha_1 \cdots \alpha_r}{}_{\beta_1 \cdots \beta_{s-1} \sigma} \,. \end{align}$

My question is, why do some terms on right-hand side of the equation have minus sign, while some have positive sign. (I notice covariant/contravariant index difference, so why do we require minus sign on some indexes?)

Tensor
  • 21
  • 1
  • I'm guessing it's because of the negative sign in the definition of covariant derivative for covariant vs contravariant tensors. When I get a minute I'll try to work it out. – levitopher Apr 14 '13 at 03:58
  • For a slightly different context and using somewhat different notation, I explained the minus sign in this answer – Yuri Vyatkin Apr 19 '13 at 07:39

0 Answers0