# Covariant divergence with Bianchi identity

The covariant derivative of some vector component $$A^\lambda$$ can be defined as $$$$∇_\mu A^\lambda = \partial_\mu {A^\lambda}+ \Gamma^\lambda_{\mu \nu} A^\nu.$$$$

Similarly, the covariant divergence can be defined as $$$$D_\mu A^\mu= ∇_\mu \cdot A^\mu = \partial_\mu {A^\mu}+ \Gamma^\mu_{\mu \lambda} A^\lambda.$$$$

Now let's consider the contracted Bianchi Identity, which says $$$$∇_\mu ( R^{\mu\nu} - \frac{1}{2}g^{\mu \nu} R) = 0.$$$$

Note that the Bianchi Identity is referring to a covariant derivative. Using this, what can we say about the divergence of this quantity? Namely, $$$$∇_\mu \cdot ( R^{\mu\nu} - \frac{1}{2}g^{\mu \nu} R) = ???$$$$

I am really tempted to say it is also equal to zero, since the first two equations seem to imply that the covariant divergence is identical to the covariant derivative when we let $$\lambda$$ = $$\mu$$ and change the dummy $$\nu$$ to a dummy $$\lambda$$. Is that correct?

For context, this is for a course for introduction to General Relativity. So assuming no torsion, assuming Ricci tensor and metric tensor are symmetric tensors, etc.

• Your second-to-last equation is already the covariant divergence - the addition of the $\cdot$ doesn’t add anything here. Perhaps you could explain what you understand the difference between the two expressions to be? Commented Dec 31, 2021 at 1:49
• I thought the second to last expression is a covariant derivative, and not a covariant divergence... unless the two are equivalent in this case? Commented Dec 31, 2021 at 1:59

Let $$T$$ be a $$(p,q)$$-tensor, so its components have $$p$$ upstairs indices and $$q$$ downstairs indices. Then the covariant derivative $$\nabla T$$ is a $$(p,q+1)$$-tensor whose components are usually written

$$\nabla_\lambda T^{\mu_1 \cdots \mu_p}_{\qquad \nu_1 \cdots \nu_q} \equiv T^{\mu_1 \cdots \mu_p}_{\qquad \nu_1 \cdots \nu_q ; \lambda}$$

For example, if $$T$$ is a $$(2,0)$$-tensor then $$\nabla T$$ would be a $$(2,1)$$-tensor with components $$\nabla_\lambda T^{\mu\nu} = \partial_\lambda T^{\mu\nu} + \Gamma^\mu_{\lambda \rho} T^{\rho \nu} + \Gamma^\nu_{\lambda \rho} T^{\mu \rho}$$

The covariant divergence of $$T$$ is what you get when you compute the covariant derivative of $$T$$ and then contract the first upstairs index with the new downstairs one. The result is a $$(p-1,q)$$-tensor with components

$$\nabla_\color{red}{\lambda} T^{\color{red}{\lambda}\cdots \mu_p}_{\qquad \nu_1 \cdots \nu_q} \equiv T^{\color{red}{\lambda} \cdots \mu_p}_{\qquad \nu_1 \cdots \nu_q ; \color{red}{\lambda}}$$ where we implicitly sum over the repeated index $$\lambda$$ as per the Einstein summation convention. If $$T$$ is again a $$(2,0)$$-tensor then $$\mathrm{div}(T)$$ would be a $$(1,0)$$-tensor (i.e. a vector field) with components

$$\nabla_\color{red}{\lambda} T^{\color{red}{\lambda}\nu} = \partial_\color{red}{\lambda} T^{\color{red}{\lambda}\nu} + \Gamma^\color{red}{\lambda}_{\color{red}{\lambda} \rho} T^{\rho \nu} + \Gamma^\nu_{\color{red}{\lambda} \rho} T^{\color{red}{\lambda} \rho}$$

• So if I understand this right, my third and fourth equations are exactly identical, because the downstairs index on the del is mu, hence why you said the dot doesn't mean anything? Commented Dec 31, 2021 at 3:00
• @Programmer Yes, that's right. The dot notation usually appears in elementary multivariable calculus when we want to distinguish the gradient of a scalar function $\nabla f$ from the divergence of a vector function $\nabla \cdot \vec F$. In index notation, $\nabla \cdot \vec F = \nabla_\mu F^\mu$. Commented Dec 31, 2021 at 3:05