[Math] Divergence in Riemannian Geometry (General Relativity)

Question

I'm taking a course in General Relativity and I'm having some problems with the notation.

I know that Einstein's tensor verifies $\nabla_aG^{ab}=0$. In physics textbooks this consequence of Bianchi identity is phrased as "the tensor has 0 divergence". I don't understand this because for me that identity means:

If you take Einstein's tensor $G$ and take the covariant derivative $\nabla_a G$ then $(\nabla_a G)^{ab}=G^{ab}_{\quad;a}=0$. I cannot see the divergence in there.

I studied that the divergence and all of those classical differential operators could be understood through the external derivative which makes sense..

So my question: Is $\nabla_a u^{a}$ called divergence because if you take the summation conventions it looks like a divergence? I come from a mathematical background and I think index notation is powerful but sometimes I feel like I'm missing the point.

Answer 1

The divergence of a vector field $V = V^a \partial _a$ on a (pseudo-)Riemannian manifold is given by $\operatorname{div} V = V^a{}_{;a}$. In words, this is obtained by taking the trace of the total covariant derivative. In the special case of $\mathbb R^n$ with a flat Euclidean or pseudo-Euclidean metric, this yields the usual calculus formula for the divergence.

By extension, it is common to define the divergence of an arbitrary tensor field as the trace of its total covariant derivative on (usually) the last two indices. So if $G$ is the (contravariant) Einstein tensor, then its divergence would be the vector field $\operatorname{div} G = G^{ba}{}_{;a} \partial_b$. Because $G$ is symmetric, this is also equal to $G^{ab}{}_{;a}\partial _b$.

You can also apply this to the covariant Einstein tensor with components $G_{ab}$; its divergence is the $1$-form $G_{ba;}{}^{a}dx^b$.