Is $\operatorname{div}(X)=\partial_i X^i$?

Question

We know that for a vector field $X$, $\operatorname{div}(X)$ is defined as $\nabla_I X^I$. This is not the same as $\partial_i X^i$ right?

I would assume that $\nabla_I X^I=\partial_i X^i-X^j\Gamma_{ij}^i $.

Answer 1

Yes, you are correct.

Another way to expand the divergence on a Riemannian manifold in local coordinates is to take the expression similar to what you find in the Laplace-Beltrami operator:

$$ X^i \partial_i \mapsto \frac{1}{\sqrt{|g|}} \partial_i \left( \sqrt{|g|} X^i \right) $$

and you see that $\nabla_I X^I$ and $\partial_i X^i$ are equal only when either the volume element relative to the local coordinate chosen has a critical point, or that $X^i$ itself vanishes at that point.