# Contracted Bianchi Identity – [PROOF] The contracted Bianchi Identity in GTR can be written as: $\nabla^\mu G_{\mu \nu}=0$
where $G_{\mu \nu}$ is the Einstein Tensor such that, $G_{\mu \nu}=R_{\mu \nu}- \frac{1}{2}g_{\mu \nu}R$
and $R_{\mu \nu}$ is the Ricci Tensor and $R$ is the Ricci Scalar.

In this post, I will be demonstrating a simple way to prove the contracted Bianchi Identity. $\nabla_{\mu}R^{\quad \enspace \delta}_{\nu \lambda \rho}+\nabla_{\nu}R^{\quad \enspace \delta}_{\lambda \mu \rho}+\nabla_{\lambda}R^{\quad \enspace \delta}_{\mu \nu \rho} = 0 \qquad (\mathrm{eq. 1})$

where $R_{\nu \lambda \rho}^{\quad \enspace \delta}$ is the Riemann Tensor,
and $\nabla_{\mu}$ is the covariant derivative.

We see that there are 5 indices in the above tensors. So we go ahead and contract alternate indices twice, to get a single index tensor.

Remember the following properties of the Riemann Tensor: $R_{\mu \nu \lambda}^{\quad \enspace \rho}=-R_{\nu \mu \lambda}^{\quad \enspace \rho}$ $R_{\mu \nu \lambda}^{\quad \enspace \rho}=-R_{\mu \nu \enspace \lambda}^{\enspace \enspace \rho}$ $R_{\mu \nu \lambda}^{\quad \enspace \rho}=R_{\lambda \enspace \mu \nu}^{\enspace \rho}$

Now, swap (interchange) the following in the first term of $\mathrm{eq1}$: $\delta$ with $\rho$, $\nu$ with $\lambda$,
and $\enspace ^\delta_\rho$ with $\lambda \nu$ $\implies \nabla_{\mu}R^{\delta}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R^{\quad \enspace \delta}_{\lambda \mu \rho}+\nabla_{\lambda}R^{\quad \enspace \delta}_{\mu \nu \rho} = 0$

[Remember Riemann Tensor is pair symmetric]

Perform a similar swapping of indices, as in the last step, on the second and third term too, to get: $\implies \nabla_{\mu}R^{\delta}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R^{\delta}_{\enspace \rho \mu \lambda}+\nabla_{\lambda}R^{\delta}_{\enspace \rho \nu \mu} = 0$

Operate $g^{\mu}_{\enspace \delta}=\delta^\mu_\delta$ on the above equation,

[Note that $g$ is covariantly constant and can be taken inside the covariant derivative] $g^{\mu}_{\enspace \delta}\left(\nabla_{\mu}R^{\delta}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R^{\delta}_{\enspace \rho \mu \lambda}+\nabla_{\lambda}R^{\delta}_{\enspace \rho \nu \mu}\right) = 0$ $\implies \nabla_{\mu}R^{\mu}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R^{\mu}_{\enspace \rho \mu \lambda}+\nabla_{\lambda}R^{\mu}_{\enspace \rho \nu \mu} = 0$ $\implies \nabla_{\mu}R^{\mu}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R_{\rho \lambda}-\nabla_{\lambda}R^{\mu}_{\enspace \rho \mu \nu} = 0$ $\implies \nabla_{\mu}R^{\mu}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R_{\rho \lambda}-\nabla_{\lambda}R_{\rho \nu} = 0$

Now, operate $g^{\nu \rho}$ on the above, $\implies g^{\nu \rho}\left(\nabla_{\mu}R^{\mu}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R_{\rho \lambda}-\nabla_{\lambda}R_{\rho \nu}\right) = 0$ $\implies \nabla_{\mu}R^{\mu}_{\enspace \lambda}+\nabla_{\nu}R_{\enspace \lambda}^{\nu}-\nabla_{\lambda}R = 0$

Since, $\nabla_{\mu}R^{\mu}_{\enspace \lambda}=\nabla^{\mu}R_{\mu \lambda}$

Therefore, $\implies \nabla^{\mu}R_{\mu \lambda}+\nabla^{\nu}R_{\nu \lambda}-\nabla_{\lambda}R = 0$

Since, $\mu$ and $\nu$ are dummy indices, therefore, we can write $\implies 2\nabla^{\mu}R_{\mu \lambda}-\nabla_{\lambda}R = 0$ $\implies 2\nabla^{\mu}R_{\mu \lambda}-g_{\mu \lambda}\nabla^{\mu}R = 0$ $\implies \nabla^{\mu} \left( R_{\mu \lambda}-\frac{g_{\mu \lambda}}{2}R \right)=0$ $\implies \boxed{\nabla^{\mu}G_{\mu \lambda}=0}$

Hence Proved.

### References: [wpedon id="7041" align="center"]