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.
Best Answer
We can show that this equality holds at any point of $M$ by computing in local coordinates. At the point $p\in M$ we use Riemannian normal coordinates, which have the property that the metric at $p$ is given by the identity matrix, $g_{ij}(p)=\delta_{ij}$, and that the derivatives of the metric coefficients are zero at $p$, as are the Christoffel symbols. With this we can write \begin{equation} \text{tr}\ \nabla X (p) = \sum_{i=1}^n g(\nabla_{\partial_i} X,\partial_i)(p) = \sum_{i=1}^n(\partial_i X^j)(p)\delta_{ij} = \sum_{i=1}^n\partial_i X^i(p). \end{equation} On the other hand, we have $\text{det}(g)(p)=1$ in these coordinates, and since the determinant is polynomial in the entries of the matrix of $g$, all its partial derivatives vanish at $p$ in these coordinates. With $dV_g = \sqrt{|g|}dx^1\wedge\ldots\wedge dx^n$ we can write \begin{equation} X\lrcorner\ dV_g = \sqrt{|g|}\sum_{i=1}^n(-1)^{i+1}X^i dx^1\wedge\ldots\widehat{dx^i}\wedge\ldots\wedge dx^n, \end{equation} where the hatted factor is the one excluded from the wedge product. Noting that the partial derivatives of $\sqrt{|g|}$ vanish at $p$ we can write \begin{equation} d(X\lrcorner\ dV_g)(p) = \sum_{i=1}^n(-1)^{i+1}(\partial_j X^i)(p)dx^j\wedge dx^1\wedge\ldots\widehat{dx^i}\wedge\ldots\wedge dx^n. \end{equation} All the terms where $i\neq j$ will be zero by antisymmetry of $\wedge$. Permuting the leftmost factor $dx^j$ to the place of the hatted factor we change sign $i-1$ times, so the sign factors cancel. We are left with \begin{equation} d(X\lrcorner\ dV_g)(p) = \left(\sum_{i=1}^n(\partial_i X^i)(p)\right)\ dx^1\wedge\ldots\wedge dx^n = \text{tr}\ \nabla X(p)\ dV_g. \end{equation}