In "Problems and Solution in Mathematics" by Ta-Tsien, 2nd Edition, exercice 3314, question b
Exercise question
For a vector field $X$ define the divergence of $X$, $\text{div}(X)$ as the trace of the operator $Y \rightarrow D_Y X$ where $D$ is the Levi-Civita connection. Find the expression for the divergence of $X$ in a local coordinate system $(x^1,…,x^n)$.
Exercise answer
Denote:
\begin{equation}
X = \sum_i X^i \frac{\partial}{\partial x^i}
\end{equation}
Then, by the definition of divergence, we have:
\begin{equation}
\text{div}(X) = \sum_{k,l} \left<D_{\frac{\partial}{\partial x^k}} X, \frac{\partial}{\partial x^l} \right> g^{kl}
=
\sum_i \left(\frac{\partial X^i}{\partial x^i} + \sum_k \Gamma^i_{ki}X^k \right)
\end{equation}
My question
I see $Y \rightarrow D_Y X$ as a "differential" operator in which for example the derivatives of the components $X^i$ and $Y^i$ of the vectors fields will appear. On the contrary, a tensor is a "local" operator which only permits algebraic operations between those components. My question thus is:
What is the coordinate expression of the tensor which trace is the divergence of $X$ ?
Best Answer
For each $Y$, $X \mapsto \nabla_Y X$ is a differential operator. But for each $X$, $Y \mapsto \nabla_Y X$ is a tensor, i.e., commutes with multiplication by functions (this is part of the definition of a connection).
We can compute the "matrix" for $Y \mapsto \nabla_Y X$ with respect to a coordinate basis $\partial_1, \dots, \partial_n$ (where $\partial_j := \frac{\partial}{\partial x^j})$: $$ \partial_j \mapsto \nabla_{\partial_j} (X^a \partial_a) = ((\partial_j X^i) + X^a \Gamma_{ja}^i ) \partial_i.$$ So the $ij$th entry of this matrix is $(\partial_j X^i) + X^a \Gamma_{ja}^i$. Said another way, the tensor in question is the $(1,1)$-tensor $$ ((\partial_j X^i) + X^a \Gamma_{ja}^i) \, \partial_i \otimes dx^j.$$ The trace of this tensor is the divergence.