TITLE: How to show that the divergence can be written as the trace of total covariant derivative, ${\rm div}={\rm tr}\circ\nabla$?
QUESTION [10 upvotes]: Let $(M,g)$ be a Riemannian manifold.

From John Lee's Riemannian Manifolds, the divergence operator $\text{div}:\Gamma(TM)\to C^{\infty}(M)$ is defined to be the map that sends $X$ to the unique function $\text{div} X$ such that
  $$(\text{div} X) \,dV_g = d(X\lrcorner \,dV_g),$$
  where $dV_g$ is the volume form and $\lrcorner$ denoted the interior multiplication. 

The above seems pretty complicated to me. Recently, I have found a much more accessible definition, i.e.
$$
\text{div} = \text{tr} \circ \nabla
$$
where $\nabla$ denotes the total covariant derivative.

How do we show that these definitions are equivalent? 

It seems to me that the books I found only use either one of these and do not mention the other.

REPLY [6 votes]: I'll add another completely different solution. If $M$ is any manifold, $\nabla$ is any torsion-free connection, $X$ is any vector field and $\beta\in \Omega^k(M)$, then $$\mathscr{L}_X\beta = \nabla_X\beta +\sum_{i=1}^k\beta(\cdot,\ldots,\nabla X,\ldots,\cdot).$$So if $\omega\in\Omega^n(M)$ is a top degree form, we have $$\mathscr{L}_X\omega = \nabla_X\omega+({\rm div}\,X)\omega$$by linear algebra. Cartan's magic formula turns that into $${\rm d}(\iota_X\omega) = \nabla_X\omega+({\rm div}\,X)\omega.$$If $\nabla$ is the Levi-Civita connection of a pseudo-Riemannian manifold $(M,g)$ and we apply this to ${\rm d}V_g$, which is parallel (as the Ricci tensor of the Levi-Civita connection is symmetric), it follows that $${\rm d}(\iota_X{\rm d}V_g)=({\rm div}\,X){\rm d}V_g$$as wanted. So this is a more powerful proof that does not rely on local computations, and holds for metrics of any signature.