Contracting \mu & \alpha - What Does It Mean?

[unscientific]

What do they mean by contracting $\mu$ with $\alpha$ ?

 

[PeterDonis]

Contraction is an operation that can be applied to any tensor or product of tensors with an upper and a lower index free. (In this case the upper index is $\mu$ and the lower index is $\alpha$.) The contraction is just a sum over all tensor components for which $\mu$ and $\alpha$ take the same value. So, for example, a tensor $T^{\mu}{}_{\alpha}$ with one upper and one lower index can be contracted to a scalar $T = T^0{}_0 + T^1{}_1 + T^2{}_2 + T^3{}_3$.

The contraction of the Bianchi identity has more terms because the identity itself has three terms, and each one becomes a sum of four terms when contracted. Note that the second step, multiplying by $g^{\nu \gamma}$, is also a contraction, because the indexes $\nu$ and $\gamma$ appear as lower indexes in the Bianchi identity.

 

[unscientific]

Contraction is an operation that can be applied to any tensor or product of tensors with an upper and a lower index free. (In this case the upper index is $\mu$ and the lower index is $\alpha$.) The contraction is just a sum over all tensor components for which $\mu$ and $\alpha$ take the same value. So, for example, a tensor $T^{\mu}{}_{\alpha}$ with one upper and one lower index can be contracted to a scalar $T = T^0{}_0 + T^1{}_1 + T^2{}_2 + T^3{}_3$.

The contraction of the Bianchi identity has more terms because the identity itself has three terms, and each one becomes a sum of four terms when contracted. Note that the second step, multiplying by $g^{\nu \gamma}$, is also a contraction, because the indexes $\nu$ and $\gamma$ appear as lower indexes in the Bianchi identity.

Do you mind showing the steps leading to the final result $2\nabla_v R_\beta^v -\nabla_\beta R = 0$?

 

[PeterDonis]

We start with

$$
\nabla_{\gamma} R^{\mu}{}_{\nu \alpha \beta} + \nabla_{\beta} R^{\mu}{}_{\nu \gamma \alpha} + \nabla_{\alpha} R^{\mu}{}_{\nu \beta \gamma} = 0
$$

Contracting $\mu$ and $\alpha$ gives (note that an index that is repeated, once upper and once lower, is summed over as I described; this is called the "Einstein summation convention" and is extremely useful):

$$
\nabla_{\gamma} R^{\alpha}{}_{\nu \alpha \beta} + \nabla_{\beta} R^{\alpha}{}_{\nu \gamma \alpha} + \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$

Completing the contractions (meaning, collapsing the contracted indexes and using the fact that contracting the Riemann tensor on the upper and the middle lower index gives the Ricci tensor) gives:

$$
\nabla_{\gamma} R_{\nu \beta} - \nabla_{\beta} R_{\nu \gamma} + \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$

where the minus sign in the second term comes in because the $\alpha$ index was the last lower index, not the middle one; swapping the indexes flips the sign (because the Riemann tensor is antisymmetric in the last two lower indexes). Now we contract with $g^{\nu \gamma}$ to give:

$$
g^{\nu \gamma} \nabla_{\gamma} R_{\nu \beta} - g^{\nu \gamma} \nabla_{\beta} R_{\nu \gamma} + g^{\nu \gamma} \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$

Completing the contractions gives (note that $R = g^{\nu \gamma} R_{\nu \gamma}$ is the Ricci scalar):

$$
\nabla^{\nu} R_{\nu \beta} - \nabla_{\beta} R + \nabla_{\alpha} R^{\alpha}{}_{\beta} = 0
$$

The first and third terms are really the same thing, because the contracted index is a "dummy" index and we can relabel it freely, and we can also freely "flip" the indexes in the contraction (to put the upper index on $R$ and the lower index on $\nabla$ in the first term). This gives what we were looking for:

$$
2 \nabla_{\nu} R^{\nu}{}_{\beta} - \nabla_{\beta} R = 0
$$