Definition of Divergence of a $2$-tensor.

Question

Lemma 7.7 in John M. Lee's book on Riemannian Geometry states the following:

Theorem. Let $M$ be a Riemannian manifold. The covariant derivatives of the Ricci and scalar curvatures satisfy $\text{div}(Ric) = \frac{1}{2}\nabla S$.

I am unable to understand the meaning of $\text{div}(Ric)$, where $\text{div}$ stands for divergence, and $Ric$ stands for the Ricci curvature. Here $S$ is the scalar curvature.

In the book the author refers to Problem 3-3 for the definition of divergence. But I checked the problem out and the divergence operator defined in the problem consumes vector fields and spits out a real number. So I am not sure what is meant by the divergence of a $2$-tensor, for instance the Ricci curvature.

Can somebody please explain the meaning of $\text{div}(Ric)$. Also, is $\nabla S$ the gradient of $S$? Thank you.

Answer 1

I found this on Wikipedia: $$\nabla_\ell R^\ell {}_m = {1 \over 2} \nabla_m R$$

Answer 2

Yes this is true, we have $$dS(X)=2div(Ric)(X)$$ Here $dS$ is the derivative of $S$ or you can write $dS(X)=g(X,grad(S))$

Also, $div(Ric)(X)= e_{i}(Ric(e_{i},X))-Ric(\nabla_{e_{i}}X,e_{i})-Rici(X,\nabla_{e_{i}} e_{i})$