Contracting R_{αβ} to R^ρ_αβσ: No 16 Multiplier

[pleasehelpmeno]

Can anyone explain how to contract $R_{\alpha \beta}$ to $R^{\rho}_{\alpha\beta\sigma}$ without multiplying it by 16 i.e $g^{\rho\xi}g_{\xi\sigma}$ It is in a sum with other tensor products and so I obviusly can't just multiply one term by anything ither than 1.

Should $\eta$'s be used although are these valid in cosmological space-times i.e $dt^2 -a^2 dx^2$

I apologise if any indices aren't in the correct order, I am self taught

 

[George Jones]

Can anyone explain how to contract $R_{\alpha \beta}$ to $R^{\rho}_{\alpha\beta\sigma}$ without multiplying it by 16 i.e $g^{\rho\xi}g_{\xi\sigma}$ It is in a sum with other tensor products and so I obviusly can't just multiply one term by anything ither than 1.

$g^{\rho\xi}g_{\xi\sigma} = \delta^\rho_\sigma$ not 16. Consequently,

$$g^{\sigma\xi}g_{\xi\rho} R^{\rho}{}_{\alpha\beta\sigma} = \delta^\sigma_\rho R^{\rho}{}_{\alpha\beta\sigma} = R^{\sigma}{}_{\alpha\beta\sigma}$$

Should $\eta$'s be used

No.